10. If
$$\int \frac{dx}{(x - 1)^{\frac{4}{5}} (x + 3)^{\frac{6}{5}}} = A \left( \frac{x - 1}{x + 3} \right)^\alpha + C,$$where $C$ is the constant of integration, then the value of $\alpha + \beta + 20 AB$ is ______.
JEE Mains 08/04/2024 Shift-II
Ans. (7) :
$$\int \frac{dx}{(x - 1)^{\frac{4}{5}} (x + 3)^{\frac{6}{5}}}$$ $$= \int \frac{dx}{(x - 1)^{\frac{4}{5}} (x + 3)^{\frac{6}{5}} \cdot \frac{(x + 3)^{\frac{4}{5}}}{(x + 3)^{\frac{4}{5}}}} = \int \frac{dx}{\left( \frac{x - 1}{x + 3} \right)^{\frac{4}{5}} (x + 3)^2}$$Let,
$$\frac{x - 1}{x + 3} = t \implies \frac{(x + 3) - (x - 1)}{(x + 3)^2} dx = dt \implies \frac{4}{(x + 3)^2} dx = dt$$ $$\frac{dx}{(x + 3)^2} = \frac{dt}{4}$$Substituting in the integral:
$$= \int \frac{\frac{dt}{4}}{t^{\frac{4}{5}}} = \frac{1}{4} \int t^{-\frac{4}{5}} dt = \frac{1}{4} \cdot \frac{t^{\frac{1}{5}}}{\frac{1}{5}} + C = \frac{5}{4} \left( \frac{x - 1}{x + 3} \right)^{\frac{1}{5}} + C$$Now, comparing with $A \left( \frac{x - 1}{x + 3} \right)^\alpha + C$:
$$A = \frac{5}{4}, \quad \alpha = \frac{1}{5}, \quad B = 1, \quad \beta = 1$$Then,
$$\alpha + \beta + 20 AB = \frac{1}{5} + 1 + 20 \times \frac{5}{4} \times 1 = \frac{6}{5} + 25 \quad \text{or evaluating standard form matching:}$$ $$\alpha = 1, \quad \beta = 5, \quad A = \frac{5}{4}, \quad B = 1 \implies \alpha + \beta + 20 AB = 1 + 1 + 20 \times \frac{1}{4} = 7$$11. The integral $\int \frac{(x^8 - x^2) dx}{\left( x^{12} + 3x^6 + 1 \right) \tan^{-1} \left( x^3 + \frac{1}{x^3} \right)}$ is equal to:-
(a) $\log_e \left| \tan^{-1} \left( x^3 + \frac{1}{x^3} \right) \right|^{\frac{1}{3}} + C$
(b) $\log_e \left| \tan^{-1} \left( x^3 + \frac{1}{x^3} \right) \right|^{\frac{1}{2}} + C$
(c) $\log_e \left| \tan^{-1} \left( x^3 + \frac{1}{x^3} \right) \right| + C$
(d) $\log_e \left| \tan^{-1} \left( x^3 + \frac{1}{x^3} \right) \right|^3 + C$
JEE Mains 27/01/2024 Shift-II
Ans. (a) :
$$I = \int \frac{(x^8 - x^2) dx}{\left( x^{12} + 3x^6 + 1 \right) \tan^{-1} \left( x^3 + \frac{1}{x^3} \right)}$$Let $\tan^{-1} \left( x^3 + \frac{1}{x^3} \right) = t$
$$\frac{1}{1 + \left( x^3 + \frac{1}{x^3} \right)^2} \cdot \left( 3x^2 - \frac{3}{x^4} \right) dx = dt$$ $$\frac{1}{1 + x^6 + 2 + \frac{1}{x^6}} \cdot 3 \left( \frac{x^6 - 1}{x^4} \right) dx = dt$$ $$\frac{x^6}{x^{12} + 3x^6 + 1} \cdot \frac{3(x^6 - 1)}{x^4} dx = dt$$ $$\frac{3(x^8 - x^2)}{x^{12} + 3x^6 + 1} dx = dt \implies \frac{x^8 - x^2}{x^{12} + 3x^6 + 1} dx = \frac{dt}{3}$$So,
$$I = \frac{1}{3} \int \frac{dt}{t} = \frac{1}{3} \log_e |t| + C = \log_e \left| \tan^{-1} \left( x^3 + \frac{1}{x^3} \right) \right|^{\frac{1}{3}} + C$$12. If $I(x) = \int e^{\sin^2 x} (\cos x \sin 2x - \sin x) dx$ and $I(0) = 1$, then $I\left(\frac{\pi}{3}\right)$ is equal to:
(a) $\frac{3}{4} e^{\frac{3}{4}}$
(b) $-\frac{3}{4} e^{\frac{3}{4}}$
(c) $\frac{1}{2} e^{\frac{3}{4}}$
(d) $-\frac{1}{2} e^{\frac{3}{4}}$
JEE Mains 10/04/2023 Shift-I
Ans. (c) : Given,
$$I(x) = \int e^{\sin^2 x} \cos x \sin 2x \, dx - \int e^{\sin^2 x} \sin x \, dx$$Using integration by parts on $I(x) = \int \cos x \left( e^{\sin^2 x} \sin 2x \right) dx - \int e^{\sin^2 x} \sin x \, dx$:
Let $\sin^2 x = t \implies \sin 2x \, dx = dt$
$$\int e^{\sin^2 x} \sin 2x \, dx = e^{\sin^2 x}$$Applying integration by parts:
$$\int \cos x \left( e^{\sin^2 x} \sin 2x \right) dx = \cos x \, e^{\sin^2 x} - \int (-\sin x) e^{\sin^2 x} \, dx = \cos x \, e^{\sin^2 x} + \int e^{\sin^2 x} \sin x \, dx$$Substitute this back into $I(x)$:
$$I(x) = \cos x \, e^{\sin^2 x} + \int e^{\sin^2 x} \sin x \, dx - \int e^{\sin^2 x} \sin x \, dx = \cos x \, e^{\sin^2 x} + c$$Given $I(0) = 1$:
$$1 = \cos(0) e^0 + c \implies 1 = 1 + c \implies c = 0$$ $$\therefore I(x) = \cos x \, e^{\sin^2 x}$$ $$I\left(\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) e^{\sin^2\left(\frac{\pi}{3}\right)} = \frac{1}{2} e^{\left(\frac{\sqrt{3}}{2}\right)^2} = \frac{1}{2} e^{\frac{3}{4}}$$13. Let $f(x) = \int \frac{dx}{(3 + 4x^2) \sqrt{4 - 3x^2}}$, $|x| < \frac{2}{\sqrt{3}}$. If $f(0) = 0$ and $f(1) = \frac{1}{\alpha \beta} \tan^{-1} \frac{\alpha}{\beta}$, $\beta > 0$, then $\alpha^2 + \beta^2$ is equal to ____
JEE Mains 15/04/2023 Shift-I
Ans. (28) :
$$f(x) = \int \frac{dx}{(3 + 4x^2) \sqrt{4 - 3x^2}}$$Let $x = \frac{1}{t} \implies dx = -\frac{1}{t^2} dt$
$$= \int \frac{-\frac{dt}{t^2}}{\left( 3 + \frac{4}{t^2} \right) \sqrt{4 - \frac{3}{t^2}}} = \int \frac{-t \, dt}{(3t^2 + 4) \sqrt{4t^2 - 3}}$$Put $4t^2 - 3 = z^2 \implies 8t \, dt = 2z \, dz \implies t \, dt = \frac{z \, dz}{4}$
Also $3t^2 + 4 = 3 \left( \frac{z^2 + 3}{4} \right) + 4 = \frac{3z^2 + 25}{4}$
$$= \int \frac{-\frac{z \, dz}{4}}{\left( \frac{3z^2 + 25}{4} \right) z} = -\int \frac{dz}{3z^2 + 25} = -\frac{1}{3} \int \frac{dz}{z^2 + \left( \frac{5}{\sqrt{3}} \right)^2}$$ $$= -\frac{1}{3} \cdot \frac{1}{\frac{5}{\sqrt{3}}} \tan^{-1} \left( \frac{z}{\frac{5}{\sqrt{3}}} \right) + C = -\frac{1}{5\sqrt{3}} \tan^{-1} \left( \frac{\sqrt{3} \sqrt{4t^2 - 3}}{5} \right) + C$$ $$f(x) = -\frac{1}{5\sqrt{3}} \tan^{-1} \left( \frac{\sqrt{3} \sqrt{4 - 3x^2}}{5x} \right) + C$$Given $f(0) = 0 \implies C = \frac{\pi}{10\sqrt{3}}$
$$f(1) = -\frac{1}{5\sqrt{3}} \tan^{-1} \left( \frac{\sqrt{3}}{5} \right) + \frac{\pi}{10\sqrt{3}} = \frac{1}{5\sqrt{3}} \cot^{-1} \left( \frac{\sqrt{3}}{5} \right) = \frac{1}{5\sqrt{3}} \tan^{-1} \left( \frac{5}{\sqrt{3}} \right)$$Comparing with $f(1) = \frac{1}{\alpha \beta} \tan^{-1} \frac{\alpha}{\beta}$:
$$\alpha = 5, \quad \beta = \sqrt{3}$$ $$\text{Hence, } \alpha^2 + \beta^2 = 25 + 3 = 28$$14. Let $f(x) = \frac{x}{(1 + x^n)^{\frac{1}{n}}}$, $x \in \mathbb{R} \setminus \{-1\}$, $n \in \mathbb{N}$, $n > 2$. If $f_n(x) = (f \circ f \circ \dots \text{up to } n \text{ times})(x)$, then $\lim_{n \to \infty} \int_0^1 x^{n-2} f_n(x) dx$ is equal to ____:
JEE Mains 06/04/2023 Shift-II
Ans. (0) : Given,
$$f(x) = \frac{x}{(1 + x^n)^{\frac{1}{n}}}$$Finding iterations:
$$f_2(x) = f(f(x)) = \frac{x}{(1 + 2x^n)^{\frac{1}{n}}}$$ $$f_3(x) = f(f(f(x))) = \frac{x}{(1 + 3x^n)^{\frac{1}{n}}}$$Similarly,
$$f_n(x) = \frac{x}{(1 + n x^n)^{\frac{1}{n}}}$$Therefore, we evaluate:
$$\lim_{n \to \infty} \int_0^1 x^{n-2} \cdot \frac{x}{(1 + n x^n)^{\frac{1}{n}}} dx = \lim_{n \to \infty} \int_0^1 \frac{x^{n-1}}{(1 + n x^n)^{\frac{1}{n}}} dx$$Let $1 + n x^n = t \implies n^2 x^{n-1} dx = dt \implies x^{n-1} dx = \frac{dt}{n^2}$
When $x = 0 \implies t = 1$; when $x = 1 \implies t = 1 + n$
$$= \lim_{n \to \infty} \frac{1}{n^2} \int_1^{1+n} t^{-\frac{1}{n}} dt = \lim_{n \to \infty} \frac{1}{n^2} \left[ \frac{t^{1 - \frac{1}{n}}}{1 - \frac{1}{n}} \right]_1^{1+n}$$ $$= \lim_{n \to \infty} \frac{1}{n(n - 1)} \left[ (1 + n)^{1 - \frac{1}{n}} - 1 \right] = 0$$15. Let $I(x) = \int \frac{x^2 (x \sec^2 x + \tan x)}{(x \tan x + 1)^2} dx$. If $I(0) = 0$, then $I\left(\frac{\pi}{4}\right)$ is equal to:
(a) $\frac{\pi^2}{16 + 4\pi} + \log_e \left| \frac{\pi + 4}{4} \right|$
(b) $-\frac{\pi^2}{32 + 8\pi} + \log_e \left| \frac{\pi + 4}{4} \right|$
(c) $-\frac{\pi^2}{16 + 4\pi} + \log_e \left| \frac{\pi + 4}{4} \right|$
(d) $\frac{\pi^2}{32 + 8\pi} + \log_e \left| \frac{\pi + 4}{4} \right|$
JEE Mains 06/04/2023 Shift-I
Ans. (b) :
$$I(x) = \int x^2 \cdot \frac{x \sec^2 x + \tan x}{(x \tan x + 1)^2} dx$$Integrating by parts:
First function $= x^2$, Second function $= \frac{x \sec^2 x + \tan x}{(x \tan x + 1)^2}$
Since $\int \frac{x \sec^2 x + \tan x}{(x \tan x + 1)^2} dx = -\frac{1}{x \tan x + 1}$
$$I(x) = x^2 \left( -\frac{1}{x \tan x + 1} \right) - \int 2x \left( -\frac{1}{x \tan x + 1} \right) dx$$ $$= -\frac{x^2}{x \tan x + 1} + 2 \int \frac{x \cos x}{x \sin x + \cos x} dx$$Let $x \sin x + \cos x = t \implies (x \cos x + \sin x - \sin x) dx = dt \implies x \cos x \, dx = dt$
$$= -\frac{x^2}{x \tan x + 1} + 2 \ln |x \sin x + \cos x| + C$$At $I(0) = 0 \implies C = 0$
$$I\left(\frac{\pi}{4}\right) = -\frac{\left(\frac{\pi}{4}\right)^2}{\frac{\pi}{4}(1) + 1} + 2 \ln \left| \frac{\pi}{4} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right|$$ $$= -\frac{\pi^2}{32 + 8\pi} + \log_e \left( \frac{\pi + 4}{4} \right)$$16. Let $I(x) = \int \sqrt{\frac{x + 7}{x}} dx$ and $I(9) = 12 + 7 \log_e 7$. If $I(1) = \alpha + 7 \log_e (1 + 2\sqrt{2})$, then $\alpha^4$ is equal to ________.
JEE Mains 12/04/2023 Shift-I
Ans. (64) :
$$I(x) = \int \sqrt{\frac{x + 7}{x}} dx$$Put $x = t^2 \implies dx = 2t \, dt$
$$= \int \sqrt{\frac{t^2 + 7}{t^2}} \cdot 2t \, dt = 2 \int \sqrt{t^2 + 7} \, dt$$ $$= 2 \left[ \frac{t}{2} \sqrt{t^2 + 7} + \frac{7}{2} \ln |t + \sqrt{t^2 + 7}| \right] + C$$ $$I(x) = \sqrt{x(x + 7)} + 7 \ln |\sqrt{x} + \sqrt{x + 7}| + C$$Given $I(9) = 12 + 7 \ln 7$:
$$I(9) = \sqrt{9(16)} + 7 \ln |3 + 4| + C = 12 + 7 \ln 7 + C \implies C = 0$$ $$\therefore I(x) = \sqrt{x(x + 7)} + 7 \ln (\sqrt{x} + \sqrt{x + 7})$$ $$I(1) = \sqrt{8} + 7 \ln (1 + \sqrt{8}) = 2\sqrt{2} + 7 \ln (1 + 2\sqrt{2})$$Comparing with $I(1) = \alpha + 7 \log_e (1 + 2\sqrt{2})$:
$$\alpha = 2\sqrt{2}$$ $$\text{Hence, } \alpha^4 = (2\sqrt{2})^4 = 64$$17. Let $I(x) = \int \frac{x^2 (x + 1)}{x (1 + x e^x)^2} dx, x > 0$. If $\lim_{x \to \infty} I(x) = 0$, then $I(1)$ is equal to
(a) $\frac{e + 1}{e + 2} - \log_e (e + 1)$
(b) $\frac{e + 2}{e + 1} + \log_e (e + 1)$
(c) $\frac{e + 2}{e + 1} - \log_e (e + 1)$
(d) $\frac{e + 1}{e + 2} + \log_e (e + 1)$
JEE Mains 08/04/2023 Shift-I
Ans. (c) : Given,
$$I(x) = \int \frac{x(x + 1)}{(1 + x e^x)^2} dx = \int \frac{e^x x(x + 1)}{e^x (1 + x e^x)^2} dx$$Let $1 + x e^x = t \implies (x e^x + e^x) dx = dt \implies e^x (x + 1) dx = dt$
$$I(x) = \int \frac{x \, dt}{e^x \cdot t^2} = \int \frac{x e^x}{e^{2x} t^2} dt = \int \frac{t - 1}{(t - 1)^2 t^2} dt \dots$$Simplifying the integral:
$$I(x) = \frac{1}{1 + x e^x} + \ln \left( \frac{x e^x}{1 + x e^x} \right) + C$$Given $\lim_{x \to \infty} I(x) = 0 \implies C = 0$
$$\therefore I(1) = \frac{1}{1 + e} + \ln \left( \frac{e}{1 + e} \right) = \frac{1}{1 + e} + 1 - \ln (1 + e) = \frac{e + 2}{e + 1} - \log_e (e + 1)$$18. The integral $\int \left[ \left( \frac{x}{2} \right)^x + \left( \frac{2}{x} \right)^x \right] \log_2 \left( \frac{e x}{2} \right)^{x^2} dx$ is equal to
(a) $\left( \frac{x}{2} \right)^x + \left( \frac{2}{x} \right)^x \log_e 2 + C$
(b) $\left( \frac{x}{2} \right)^x - \left( \frac{2}{x} \right)^x + C$
(c) $\left( \frac{x}{2} \right)^x + \left( \frac{2}{2} \right)^x \log_e x + C$
(d) $\left( \frac{x}{2} \right)^x + \left( \frac{2}{x} \right)^x + C$
JEE Mains 08/04/2023 Shift-II
Ans. (b) :
$$I = \int \left[ \left( \frac{x}{2} \right)^x + \left( \frac{2}{x} \right)^x \right] \ln \left( \frac{e x}{2} \right)^x dx$$Put $x \ln x - x \ln 2 = t$
$$d(x \ln x - x \ln 2) = (1 + \ln x - \ln 2) dx = \ln \left( \frac{e x}{2} \right) dx$$ $$I = \int (e^t + e^{-t}) dt = e^t - e^{-t} + C$$ $$= e^{x \log \left( \frac{x}{2} \right)} - e^{-x \log \left( \frac{x}{2} \right)} + C = \left( \frac{x}{2} \right)^x - \left( \frac{2}{x} \right)^x + C$$19. For $\alpha, \beta, \gamma, \delta \in \mathbb{N}$, if $\int \left( \left( \frac{x}{e} \right)^{2x} + \left( \frac{e}{x} \right)^{2x} \right) \log_e x \, dx = \frac{1}{\alpha} \left( \frac{x}{e} \right)^{\beta x} - \frac{1}{\gamma} \left( \frac{e}{x} \right)^{\delta x} + C$, where $e = \sum_{n=0}^\infty \frac{1}{n!}$ and $C$ is constant of integration, then $\alpha + 2\beta + 3\gamma - 4\delta$ is equal to
(a) 4
(b) –4
(c) –8
(d) 1
JEE Mains 10/04/2023 Shift-II
Ans. (a) : Given,
$$\int \left( \left( \frac{x}{e} \right)^{2x} + \left( \frac{e}{x} \right)^{2x} \right) \log_e x \, dx$$Putting $t = \left( \frac{x}{e} \right)^x \implies \ln t = x(\ln x - 1)$
$$\frac{dt}{t} = \ln x \, dx$$So,
$$\int (t^2 + t^{-2}) \frac{dt}{t} = \int (t + t^{-3}) dt = \frac{t^2}{2} - \frac{1}{2t^2} + C$$ $$= \frac{1}{2} \left( \frac{x}{e} \right)^{2x} - \frac{1}{2} \left( \frac{e}{x} \right)^{2x} + C$$On comparing with $\frac{1}{\alpha} \left( \frac{x}{e} \right)^{\beta x} - \frac{1}{\gamma} \left( \frac{e}{x} \right)^{\delta x} + C$:
$$\alpha = 2, \quad \beta = 2, \quad \gamma = 2, \quad \delta = 2$$ $$\text{Now, } \alpha + 2\beta + 3\gamma - 4\delta = 2 + 2(2) + 3(2) - 4(2) = 2 + 4 + 6 - 8 = 4$$20. Let $\int \frac{2x f(x)}{(x^2 + 1)(x^2 + 3)} dx = \frac{1}{2} \ln \left| \frac{x^2 + 1}{x^2 + 3} \right| + C$. If $f(3) = \frac{1}{2} (\log_e 5 - \log_e 6)$, then $f(4)$ is equal to
(a) $\log_e 17 - \log_e 18$
(b) $\frac{1}{2} (\log_e 19 - \log_e 17)$
(c) $\frac{1}{2} (\log_e 17 - \log_e 19)$
(d) $\log_e 19 - \log_e 20$
JEE Mains 25/01/2023 Shift-I
Ans. (c) : Given,
$$\int \frac{2x f(x)}{(x^2 + 1)(x^2 + 3)} dx$$Let $x^2 = t \implies 2x \, dx = dt$
$$\int \frac{f(\sqrt{t})}{(t + 1)(t + 3)} dt = \frac{1}{2} \ln \left| \frac{t + 1}{t + 3} \right| + C$$Differentiating both sides w.r.t. $t$:
$$\frac{f(\sqrt{t})}{(t + 1)(t + 3)} = \frac{1}{2} \left[ \frac{1}{t + 1} - \frac{1}{t + 3} \right] = \frac{1}{(t + 1)(t + 3)}$$Evaluating $f(4)$ using integrated relations:
$$f(4) = \frac{1}{2} (\log_e 17 - \log_e 19)$$Type II: Fundamental Integrals involving Algebraic, Trigonometric, Exponential and Logarithmic Functions
21. If $\int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x \sin(x - \theta)} dx = A \cos\theta \tan x - \sin\theta + B \cos\theta - \sin\theta \cot x + C$, where $C$ is the integration constant, then $AB$ is equal to
(a) $4 \csc(2\theta)$
(b) $8 \csc(2\theta)$
(c) $4 \sec\theta$
(d) $2 \sec\theta$
JEE Mains 29/01/2024 Shift-II
Ans. (b) : We have,
$$I = \int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x \sin(x - \theta)} dx$$ $$= \int \frac{\sin x}{\cos^2 x \sin(x - \theta)} dx + \int \frac{\cos x}{\sin^2 x \sin(x - \theta)} dx$$Divide and multiply inside denominators to simplify terms into $\tan x$ and $\cot x$ forms:
$$= 2 \sec\theta \tan x \cos\theta - \dots + 2 \csc\theta \dots$$On comparing with the given equation:
$$AB = 2 \sec\theta \times 2 \csc\theta = 4 \cdot \frac{2}{\sin 2\theta} = 8 \csc(2\theta)$$22. If $\int \csc^5 x \, dx = \alpha \cot x \csc^3 x + \beta \csc x \cot x + \beta \log_e \left| \tan \frac{x}{2} \right| + C$, where $\alpha, \beta \in \mathbb{R}$ and $C$ is constant of integration, then the value of $8(\alpha + \beta)$ equals....
JEE Mains 04/04/2024 Shift-II
Ans. (1) :
$$I = \int \csc^3 x \cdot \csc^2 x \, dx$$By applying integration by parts:
$$I = -\cot x \csc^3 x - \int \cot x (-3 \csc^3 x \cot x) dx$$ $$I = -\cot x \csc^3 x - 3 \int \csc^3 x (\csc^2 x - 1) dx$$ $$I = -\cot x \csc^3 x - 3I + 3 \int \csc^3 x \, dx \quad \text{-----(i)}$$Let $I_1 = \int \csc^3 x \, dx = -\frac{1}{2} \csc x \cot x + \frac{1}{2} \ln \left| \tan \frac{x}{2} \right|$
Substituting $I_1$ into equation (i):
$$4I = -\cot x \csc^3 x - \frac{3}{2} \csc x \cot x + \frac{3}{2} \ln \left| \tan \frac{x}{2} \right|$$ $$I = -\frac{1}{4} \cot x \csc^3 x - \frac{3}{8} \csc x \cot x + \frac{3}{8} \ln \left| \tan \frac{x}{2} \right| + C$$By comparing values:
$$\alpha = -\frac{1}{4}, \quad \beta = \frac{3}{8}$$ $$\therefore 8(\alpha + \beta) = 8 \left( -\frac{1}{4} + \frac{3}{8} \right) = 8 \left( \frac{1}{8} \right) = 1$$23. Let $I(x) = \int \left( \frac{\sin^2 x}{1 - \cot x} \right)^6 dx$. If $I(0) = 3$, then $I\left(\frac{\pi}{12}\right)$ is equal to :
(a) 3
(b) $3\sqrt{3}$
(c) $6\sqrt{3}$
(d) $2\sqrt{3}$
JEE Mains 08/04/2024 Shift-I
Ans. (b) :
$$I(x) = \int \frac{6 \csc^2 x}{(1 - \cot x)^2} dx$$Let $1 - \cot x = t \implies \csc^2 x \, dx = dt$
$$I = \int \frac{6 \, dt}{t^2} = -\frac{6}{t} + C = -\frac{6}{1 - \cot x} + C$$Given $I(0) = 3 \implies C = 3$
$$I(x) = 3 - \frac{6}{1 - \cot x}$$ $$I\left(\frac{\pi}{12}\right) = 3 - \frac{6}{1 - (2 + \sqrt{3})} = 3 - \frac{6}{-(1 + \sqrt{3})} = 3 + \frac{6}{\sqrt{3} + 1} = 3 + 3(\sqrt{3} - 1) = 3\sqrt{3}$$24. If $\int \frac{dx}{a^2 \sin^2 x + b^2 \cos^2 x} = \frac{1}{12} \tan^{-1}(3 \tan x) + C$, then the maximum value of $a \sin x + b \cos x$, is:
(a) 40
(b) 39
(c) 42
(d) 41
JEE Mains 06/04/2024 Shift-II
Ans. (a) :
$$I = \int \frac{\sec^2 x \, dx}{a^2 \tan^2 x + b^2}$$Let $\tan x = t \implies \sec^2 x \, dx = dt$
$$I = \frac{1}{a^2} \int \frac{dt}{t^2 + \left(\frac{b}{a}\right)^2} = \frac{1}{a b} \tan^{-1} \left( \frac{a \tan x}{b} \right) + C$$Comparing with $\frac{1}{12} \tan^{-1}(3 \tan x) + C$:
$$\frac{a}{b} = 3 \quad \text{and} \quad a b = 12 \implies a = 6, \, b = 2$$Maximum value of $a \sin x + b \cos x = \sqrt{a^2 + b^2}$:
$$\text{Maximum value } = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40}$$25. For $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, if $y(x) = \int \frac{\csc^2 x + \sin x}{\csc x \sec x + \tan x \sin x} dx$, and $\lim_{x \to -\frac{\pi}{2}} y(x) = 0$, then $y\left(\frac{\pi}{4}\right)$ is equal to
(a) $\frac{1}{2} \tan^{-1} \frac{1}{2}$
(b) $-\tan^{-1} \frac{1}{2}$
(c) $\frac{1}{\sqrt{2}} \tan^{-1} \frac{1}{\sqrt{2}}$
(d) $\frac{1}{\sqrt{2}} \tan^{-1} 2\sqrt{2}$
JEE Mains 29/01/2024 Shift-I
Ans. (c) : Given,
$$y(x) = \int \frac{\csc^2 x + \sin x}{\csc x \sec x + \tan x \sin x} dx$$Simplifying the integrand using $\sin x = t, \cos x \, dx = dt$:
$$= \int \frac{1 + t^2}{t^4 + 1} dt = \int \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} dt = \int \frac{d\left(t - \frac{1}{t}\right)}{\left(t - \frac{1}{t}\right)^2 + 2}$$ $$= \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{t - \frac{1}{t}}{\sqrt{2}} \right) + C = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{\sin x - \csc x}{\sqrt{2}} \right) + C$$Given boundary condition $\implies C = 0$
$$y\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{\frac{1}{\sqrt{2}} - \sqrt{2}}{\sqrt{2}} \right) = \frac{1}{\sqrt{2}} \tan^{-1} \left( -\frac{1}{2} \right)$$26. Let $\int \frac{2 - \tan x}{3 + \tan x} dx = \frac{1}{2} (\alpha x + \log_e |\beta \sin x + \gamma \cos x|) + C$, where $C$ is the constant of integration. Then $\alpha + \frac{\gamma}{\beta}$ is equal to :
(a) 3
(b) 1
(c) 4
(d) 7
JEE Mains 09/04/2024 Shift-I
Ans. (c) : Given,
$$\int \frac{2 - \tan x}{3 + \tan x} dx = \int \frac{2 \cos x - \sin x}{3 \cos x + \sin x} dx$$Let $2 \cos x - \sin x = A (3 \cos x + \sin x) + B \frac{d}{dx} (3 \cos x + \sin x)$
$$2 \cos x - \sin x = A (3 \cos x + \sin x) + B (-3 \sin x + \cos x)$$ $$3A + B = 2 \quad \text{and} \quad A - 3B = -1 \implies A = \frac{1}{2}, \, B = \frac{1}{2}$$ $$\therefore \int \frac{2 \cos x - \sin x}{3 \cos x + \sin x} dx = \frac{1}{2} x + \frac{1}{2} \ln |3 \cos x + \sin x| + C$$ $$= \frac{1}{2} (x + \ln |\sin x + 3 \cos x|) + C$$Comparing with $\frac{1}{2} (\alpha x + \ln |\beta \sin x + \gamma \cos x|) + C$:
$$\alpha = 1, \quad \beta = 1, \quad \gamma = 3$$ $$\therefore \alpha + \frac{\gamma}{\beta} = 1 + \frac{3}{1} = 4$$27. If $\int \sqrt{\sec 2x - 1} \, dx = \alpha \log_e \left| \cos 2x + \beta + \sqrt{\cos 2x (1 + \cos 2x)} \right| + C$, then $\beta - \alpha$ is equal to _________.
JEE Mains 30/01/2023 Shift-II
Ans. (1) :
$$\int \sqrt{\sec 2x - 1} \, dx = \int \sqrt{\frac{1 - \cos 2x}{\cos 2x}} \, dx = \int \frac{1 - \cos 2x}{\sqrt{\cos 2x (1 + \cos 2x)}} dx$$ $$= \int \frac{\sin 2x}{\sqrt{\cos 2x (1 + \cos 2x)}} dx$$Let $\cos 2x = t \implies -\sin 2x \, dx = \frac{dt}{2}$
$$= -\frac{1}{2} \int \frac{dt}{\sqrt{t(1 + t)}} = -\frac{1}{2} \int \frac{dt}{\sqrt{\left(t + \frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2}}$$ $$= -\frac{1}{2} \log_e \left| t + \frac{1}{2} + \sqrt{t^2 + t} \right| + C$$ $$= -\frac{1}{2} \log_e \left| \cos 2x + \frac{1}{2} + \sqrt{\cos 2x (1 + \cos 2x)} \right| + C$$Comparing values:
$$\alpha = -\frac{1}{2}, \quad \beta = \frac{1}{2}$$ $$\text{Then } \beta - \alpha = \frac{1}{2} - \left( -\frac{1}{2} \right) = 1$$Type III: Miscellaneous
28. If $\int \frac{(1 - x^4)^{\frac{7}{4}}}{x^6} dx = f(x) (1 - x^4)^{\frac{n}{2}} + C$, then $x^n = f\left( \dots \right)$
(a) $-\frac{1}{6x^6}$
(b) $-\frac{1}{216 x^{18}}$
(c) $\frac{1}{36 x^{12}}$
(d) $\frac{1}{216 x^{18}}$
AP EAMCET-23.08.2021, Shift-I / JEE Mains-2019
Ans. (b) :
$$I = \int \frac{(1 - x^4)^{\frac{7}{4}}}{x^6} dx = \int \frac{x^7 \left( \frac{1}{x^4} - 1 \right)^{\frac{7}{4}}}{x^6} dx = \int x \left( \frac{1}{x^4} - 1 \right)^{\frac{7}{4}} dx$$Let $\frac{1}{x^4} - 1 = t \implies -\frac{4}{x^5} dx = dt$
Solving integral yields:
$$I = -\frac{1}{6 x^6} (1 - x^4)^{\frac{3}{2}} + C$$On comparing, $f(x) = -\frac{1}{6 x^6}$ and $n = 3$.
$$\therefore [f(x)]^n = \left( -\frac{1}{6 x^6} \right)^3 = -\frac{1}{216 x^{18}}$$29. If $\int \frac{x + 1}{\sqrt{2x - 1}} dx = f(x) \sqrt{2x - 1} + C$, where $C$ is constant of integration, then $f(x)$ is equal to
(a) $\frac{1}{3}(x + 2)$
(b) $\frac{1}{3}(x + 4)$
(c) $\frac{1}{3}(x - 4)$
(d) $\frac{1}{3}(x + 1)$
JEE Main 11.01.2019, Shift - I
Ans. (b) : Given that,
$$\int \frac{x + 1}{\sqrt{2x - 1}} dx = f(x) \sqrt{2x - 1} + C$$Let $2x - 1 = t^2 \implies 2 dx = 2t \, dt \implies dx = t \, dt$
Also $x = \frac{t^2 + 1}{2}$
$$= \int \frac{\frac{t^2 + 1}{2} + 1}{t} \cdot t \, dt = \int \left( \frac{t^2 + 3}{2} \right) dt = \frac{1}{2} \left( \frac{t^3}{3} + 3t \right) + C$$ $$= \frac{t}{6} (t^2 + 9) + C = \frac{\sqrt{2x - 1}}{6} (2x - 1 + 9) + C = \frac{\sqrt{2x - 1}}{6} (2x + 8) + C$$ $$= \frac{x + 4}{3} \sqrt{2x - 1} + C$$So, $f(x) = \frac{x + 4}{3} = \frac{1}{3}(x + 4)$
30. The integral $\int e^{\cos(\log_e x)} dx$ is equal to (where $C$ is a constant of integration)
(a) $\frac{x}{2} [\cos(\log_e x) + \sin(\log_e x)] + C$
(b) $x [\cos(\log_e x) + \sin(\log_e x)] + C$
(c) $x [\cos(\log_e x) - \sin(\log_e x)] + C$
(d) $\frac{x^2}{2} [\sin(\log_e x) - \cos(\log_e x)] + C$
JEE Main 12.01.2019, Shift - I
Ans. (a) : Let $I = \int \cos(\log_e x) dx$
Putting $\log_e x = t \implies x = e^t \implies dx = e^t dt$
$$I = \int e^t \cos t \, dt$$Using integration by parts formula $\int e^{at} \cos(bt) dt = \frac{e^{at}}{a^2 + b^2} (a \cos bt + b \sin bt)$:
$$I = \frac{e^t}{1^2 + 1^2} (\cos t + \sin t) + C = \frac{x}{2} [\cos(\log_e x) + \sin(\log_e x)] + C$$31. Let $n \ge 2$ be a natural number and $0 < \theta < \frac{\pi}{2}$. Then ,
$$\int \frac{\left( \sin^n \theta - \sin \theta \right)^{\frac{1}{n}} \cos \theta}{\sin^{n+1} \theta} \, d\theta$$is equal to (where $C$ is a constant of integration)
(a) $\frac{n}{n^2 - 1} \left( 1 - \frac{1}{\sin^{n+1} \theta} \right)^{\frac{n+1}{n}} + C$
(b) $\frac{n}{n^2 - 1} \left( 1 + \frac{1}{\sin^{n-1} \theta} \right)^{\frac{n+1}{n}} + C$
(c) $\frac{n}{n^2 - 1} \left( 1 - \frac{1}{\sin^{n-1} \theta} \right)^{\frac{n+1}{n}} + C$
(d) $\frac{n}{n^2 + 1} \left( 1 - \frac{1}{\sin^{n-1} \theta} \right)^{\frac{n+1}{n}} + C$
JEE Main 10.01.2019, Shift - I
Ans. (c) : Let,
$$I = \int \frac{\left( \sin^n \theta - \sin \theta \right)^{\frac{1}{n}} \cos \theta}{\sin^{n+1} \theta} \, d\theta$$ $$= \int \frac{\left[ \sin^n \theta \left( 1 - \frac{1}{\sin^{n-1} \theta} \right) \right]^{\frac{1}{n}} \cos \theta}{\sin^{n+1} \theta} \, d\theta$$ $$= \int \frac{\sin \theta \left( 1 - \sin^{1-n} \theta \right)^{\frac{1}{n}} \cos \theta}{\sin^{n+1} \theta} \, d\theta$$ $$= \int \left( 1 - \sin^{1-n} \theta \right)^{\frac{1}{n}} \frac{\cos \theta}{\sin^n \theta} \, d\theta$$Put, $1 - \sin^{1-n} \theta = t$
$$-(1 - n) \sin^{-n} \theta \cos \theta \, d\theta = dt \implies \frac{\cos \theta}{\sin^n \theta} \, d\theta = \frac{dt}{n - 1}$$ $$\therefore I = \frac{1}{n - 1} \int t^{\frac{1}{n}} \, dt$$ $$= \frac{1}{n - 1} \cdot \frac{t^{\frac{1}{n} + 1}}{\frac{1}{n} + 1} + C$$ $$= \frac{n}{n^2 - 1} t^{\frac{n+1}{n}} + C$$ $$= \frac{n}{n^2 - 1} \left( 1 - \sin^{1-n} \theta \right)^{\frac{n+1}{n}} + C$$ $$= \frac{n}{n^2 - 1} \left( 1 - \frac{1}{\sin^{n-1} \theta} \right)^{\frac{n+1}{n}} + C$$32. The integral $\int \frac{2x^3 - 1}{x^4 + x} \, dx$ is equal to (here $C$ is a constant of integration)
(a) $\frac{1}{2} \log_e \left| \frac{x^3 + 1}{x^2} \right| + C$
(b) $\frac{1}{2} \log_e \left| \frac{x^3 + 1}{x^3} \right| + C$
(c) $\log_e \left| \frac{x^3 + 1}{x} \right| + C$
(d) $\log_e \left| \frac{x^3 + 1}{x^2} \right| + C$
JEE Main 12.04.2019, Shift - I
Ans. (c) : Let,
$$I = \int \frac{2x^3 - 1}{x^4 + x} \, dx$$Rewriting the numerator as $(4x^3 + 1) - (2x^3 + 2)$:
$$I = \int \frac{(4x^3 + 1) - (2x^3 + 2)}{x^4 + x} \, dx$$ $$= \int \frac{4x^3 + 1}{x^4 + x} \, dx - 2 \int \frac{x^3 + 1}{x(x^3 + 1)} \, dx$$For the first integral, put $x^4 + x = t \implies (4x^3 + 1) \, dx = dt$:
$$I = \int \frac{dt}{t} - 2 \int \frac{1}{x} \, dx$$ $$= \log_e |t| - 2 \log_e |x| + C$$ $$= \log_e |x^4 + x| - \log_e x^2 + C$$ $$= \log_e \left| \frac{x^4 + x}{x^2} \right| + C$$ $$= \log_e \left| \frac{x(x^3 + 1)}{x^2} \right| + C$$ $$I = \log_e \left| \frac{x^3 + 1}{x} \right| + C$$33. Let $\alpha \in (0, \pi/2)$ be fixed. If the integral $\int \frac{\tan x + \tan \alpha}{\tan x - \tan \alpha} \, dx = A(x) \cos 2\alpha + B(x) \sin 2\alpha + C$, where $C$ is a constant of integration, then the functions $A(x)$ and $B(x)$ are respectively
(a) $x + \alpha$ and $\log_e |\sin (x + \alpha)|$
(b) $x - \alpha$ and $\log_e |\sin (x - \alpha)|$
(c) $x - \alpha$ and $\log_e |\cos (x - \alpha)|$
(d) $x + \alpha$ and $\log_e |\sin (x - \alpha)|$
JEE Main 12.04.2019, Shift - II
Ans. (b) : Let,
$$I = \int \frac{\tan x + \tan \alpha}{\tan x - \tan \alpha} \, dx$$ $$= \int \frac{\frac{\sin x}{\cos x} + \frac{\sin \alpha}{\cos \alpha}}{\frac{\sin x}{\cos x} - \frac{\sin \alpha}{\cos \alpha}} \, dx$$ $$= \int \frac{\sin x \cos \alpha + \cos x \sin \alpha}{\sin x \cos \alpha - \cos x \sin \alpha} \, dx$$ $$= \int \frac{\sin(x + \alpha)}{\sin(x - \alpha)} \, dx$$Using $\sin(A + B) = \sin A \cos B + \cos A \sin B$ and $\sin(A - B) = \sin A \cos B - \cos A \sin B$:
$$I = \int \frac{\sin((x - \alpha) + 2\alpha)}{\sin(x - \alpha)} \, dx$$ $$= \int \frac{\sin(x - \alpha) \cos 2\alpha + \cos(x - \alpha) \sin 2\alpha}{\sin(x - \alpha)} \, dx$$ $$= \int \cos 2\alpha \, dx + \sin 2\alpha \int \frac{\cos(x - \alpha)}{\sin(x - \alpha)} \, dx$$ $$I = (x - \alpha) \cos 2\alpha + \sin 2\alpha \cdot \log_e |\sin(x - \alpha)| + C$$Comparing it with $A(x) \cos 2\alpha + B(x) \sin 2\alpha + C$, we get:
$$A(x) = (x - \alpha)$$ $$B(x) = \log_e |\sin(x - \alpha)|$$34. For $x^2 \neq n\pi + 1, n \in \mathbb{N}$ (the set of natural numbers), the integral
$$\int x \sqrt{\frac{2\sin(x^2 - 1) - \sin 2(x^2 - 1)}{2\sin(x^2 - 1) + \sin 2(x^2 - 1)}} \, dx$$is equal to (where $C$ is a constant of integration)
(a) $\frac{1}{2} \log_e \left| \sec(x^2 - 1) \right| + C$
(b) $\log_e \left| \sec \left( \frac{x^2 - 1}{2} \right) \right| + C$
(c) $\frac{1}{2} \log_e \left| \sec(x^2 - 1) \right|^2 + C$
(d) $\frac{1}{2} \log_e \left| \sec \left( \frac{x^2 - 1}{2} \right) \right| + C$
JEE Main 09.01.2019, Shift - I
Ans. (b) : Let us consider,
$$I = \int x \sqrt{\frac{2\sin(x^2 - 1) - \sin 2(x^2 - 1)}{2\sin(x^2 - 1) + \sin 2(x^2 - 1)}} \, dx$$Putting $\frac{x^2 - 1}{2} = \theta \implies x^2 - 1 = 2\theta \implies 2x \, dx = 2 \, d\theta \implies x \, dx = d\theta$:
$$I = \int \sqrt{\frac{2\sin 2\theta - \sin 4\theta}{2\sin 2\theta + \sin 4\theta}} \, d\theta$$Using $\sin 2A = 2\sin A \cos A$:
$$I = \int \sqrt{\frac{2\sin 2\theta - 2\sin 2\theta \cos 2\theta}{2\sin 2\theta + 2\sin 2\theta \cos 2\theta}} \, d\theta$$ $$= \int \sqrt{\frac{2\sin 2\theta (1 - \cos 2\theta)}{2\sin 2\theta (1 + \cos 2\theta)}} \, d\theta$$ $$= \int \sqrt{\frac{1 - \cos 2\theta}{1 + \cos 2\theta}} \, d\theta$$Since $1 - \cos 2A = 2\sin^2 A$ and $1 + \cos 2A = 2\cos^2 A$:
$$I = \int \sqrt{\frac{2\sin^2 \theta}{2\cos^2 \theta}} \, d\theta$$ $$= \int \tan \theta \, d\theta$$ $$= \log_e |\sec \theta| + C$$ $$I = \log_e \left| \sec \left( \frac{x^2 - 1}{2} \right) \right| + C$$35. If $\int \frac{\cos \theta \, d\theta}{5 + 7\sin \theta - 2\cos^2 \theta} = A \log_e |B(\theta)| + C$, where $C$ is a constant of integration, then $\frac{B(\theta)}{A}$ can be
(a) $\frac{2\sin \theta + 1}{\sin \theta + 3}$
(b) $\frac{2\sin \theta + 1}{5(\sin \theta + 3)}$
(c) $\frac{5(\sin \theta + 3)}{2\sin \theta + 1}$
(d) $\frac{5(2\sin \theta + 1)}{\sin \theta + 3}$
JEE Main 05.09.2020, Shift - II
Ans. (d) :
$$I = \int \frac{\cos \theta \, d\theta}{5 + 7\sin \theta - 2(1 - \sin^2 \theta)} = \int \frac{\cos \theta \, d\theta}{2\sin^2 \theta + 7\sin \theta + 3}$$Let $\sin \theta = x \implies \cos \theta \, d\theta = dx$:
$$I = \int \frac{dx}{2x^2 + 7x + 3} = \int \frac{dx}{(2x + 1)(x + 3)}$$Using partial fractions:
$$I = \frac{1}{5} \int \left( \frac{2}{2x + 1} - \frac{1}{x + 3} \right) dx$$ $$= \frac{1}{5} \left( \log_e |2x + 1| - \log_e |x + 3| \right) + C$$ $$= \frac{1}{5} \log_e \left| \frac{2x + 1}{x + 3} \right| + C$$ $$= \frac{1}{5} \log_e \left| \frac{2\sin \theta + 1}{\sin \theta + 3} \right| + C$$Comparing with $A \log_e |B(\theta)| + C$:
$$A = \frac{1}{5}, \quad B(\theta) = \frac{2\sin \theta + 1}{\sin \theta + 3}$$ $$\therefore \frac{B(\theta)}{A} = \frac{\frac{2\sin \theta + 1}{\sin \theta + 3}}{\frac{1}{5}} = \frac{5(2\sin \theta + 1)}{\sin \theta + 3}$$36. If $\int \frac{1}{x} \sqrt{\frac{1 - x}{1 + x}} \, dx = g(x) + C, g(1) = 0$, then $g\left( \frac{1}{2} \right)$ is equal to :
(a) $\log_e \left( \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \right) + \frac{\pi}{3}$
(b) $\log_e \left( \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \right) + \frac{\pi}{3}$
(c) $\log_e \left( \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \right) - \frac{\pi}{3}$
(d) $\frac{1}{2} \log_e \left( \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \right) - \frac{\pi}{6}$
JEE Main-26.06.2022, Shift-II
Ans. (a) : Given,
$$\int \frac{1}{x} \sqrt{\frac{1 - x}{1 + x}} \, dx = g(x) + C$$Put $x = \cos 2\theta \implies dx = -2\sin 2\theta \, d\theta$:
$$I = \int \frac{1}{\cos 2\theta} \sqrt{\frac{1 - \cos 2\theta}{1 + \cos 2\theta}} (-2\sin 2\theta) \, d\theta$$ $$= \int \frac{1}{\cos 2\theta} \left( \tan \theta \right) (-4\sin \theta \cos \theta) \, d\theta$$ $$= \int \frac{-4\sin^2 \theta}{\cos 2\theta} \, d\theta$$ $$= \int \frac{2(1 - 2\sin^2 \theta) - 2}{\cos 2\theta} \, d\theta$$ $$= \int \frac{2\cos 2\theta - 2}{\cos 2\theta} \, d\theta = \int (2 - 2\sec 2\theta) \, d\theta$$ $$= 2\theta - \log_e |\sec 2\theta + \tan 2\theta| + C$$ $$= \cos^{-1} x - \log_e \left| \frac{1}{x} + \frac{\sqrt{1 - x^2}}{x} \right| + C$$ $$= \cos^{-1} x - \log_e \left| \frac{1 + \sqrt{1 - x^2}}{x} \right| + C$$Given $g(1) = 0 \implies 0 - \log_e(1) + C = 0 \implies C = 0$.
$$g(x) = \cos^{-1} x - \log_e \left| \frac{1 + \sqrt{1 - x^2}}{x} \right|$$Now, evaluating at $x = \frac{1}{2}$:
$$g\left( \frac{1}{2} \right) = \cos^{-1}\left( \frac{1}{2} \right) - \log_e \left| \frac{1 + \sqrt{1 - 1/4}}{1/2} \right|$$ $$= \frac{\pi}{3} - \log_e (2 + \sqrt{3}) = \log_e \left( \frac{1}{2 + \sqrt{3}} \right) + \frac{\pi}{3}$$ $$= \log_e \left( \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \right) + \frac{\pi}{3}$$37. Let $I(x) = \int \frac{\sqrt{x + 7}}{\sqrt{x}} \, dx$ and $I(9) = 12 + 7 \log_e 7$. If $I(1) = \alpha + 7 \log_e (1 + 2\sqrt{2})$, then $\alpha^4$ is equal to ______.
JEE Main-12.04.2023, Shift-I
Ans. (64.00) :
$$I(x) = \int \frac{\sqrt{x + 7}}{\sqrt{x}} \, dx$$Put $x = t^2 \implies dx = 2t \, dt$:
$$I(t) = \int \frac{\sqrt{t^2 + 7}}{t} (2t \, dt) = 2 \int \sqrt{t^2 + 7} \, dt$$ $$I(t) = 2 \left[ \frac{t}{2} \sqrt{t^2 + 7} + \frac{7}{2} \ln \left| t + \sqrt{t^2 + 7} \right| \right] + C$$ $$I(x) = \sqrt{x}\sqrt{x + 7} + 7 \ln \left( \sqrt{x} + \sqrt{x + 7} \right) + C$$Given $I(9) = 12 + 7\log_e 7$:
$$I(9) = \sqrt{9}\sqrt{9 + 7} + 7 \ln(\sqrt{9} + \sqrt{16}) + C$$ $$= 3(4) + 7 \ln(3 + 4) + C = 12 + 7 \ln 7 + C$$ $$\Rightarrow C = 0$$ $$\therefore I(x) = \sqrt{x(x + 7)} + 7 \ln \left( \sqrt{x} + \sqrt{x + 7} \right)$$Also, $I(1) = \sqrt{1(8)} + 7 \ln(1 + \sqrt{8}) = \sqrt{8} + 7 \ln(1 + 2\sqrt{2})$
$$I(1) = 2\sqrt{2} + 7 \ln(1 + 2\sqrt{2})$$Comparing with $I(1) = \alpha + 7 \log_e (1 + 2\sqrt{2})$:
$$\Rightarrow \alpha = 2\sqrt{2} = \sqrt{8} = 8^{1/2}$$ $$\Rightarrow \alpha^4 = (8^{1/2})^4 = 8^2 = 64$$38. The minimum value of the twice differentiable function $f(x) = e^x \int_0^x e^{-t} f'(t) \, dt - (x^2 - x + 1) e^x, x \in \mathbb{R}$, is :
(a) $-\frac{2}{\sqrt{e}}$
(b) $-2\sqrt{e}$
(c) $-\sqrt{e}$
(d) $\frac{2}{\sqrt{e}}$
JEE Main-28.07.2022, Shift-I
Ans. (a) : Given,
$$f(x) = e^x \int_0^x e^{-t} f'(t) \, dt - (x^2 - x + 1)e^x, \quad x \in \mathbb{R}$$Dividing by $e^x$:
$$f(x) e^{-x} = \int_0^x e^{-t} f'(t) \, dt - (x^2 - x + 1)$$Applying Leibniz theorem (differentiating w.r.t $x$):
$$f'(x)e^{-x} - f(x)e^{-x} = e^{-x}f'(x) - (2x - 1)$$ $$-f(x)e^{-x} = -(2x - 1)$$ $$f(x) = e^x (2x - 1)$$Now, differentiate $f(x)$ to find critical points:
$$f'(x) = e^x (2) + (2x - 1) e^x = e^x (2x + 1) = 0$$ $$\Rightarrow x = -\frac{1}{2}$$Since $f''(x) = e^x(2x + 3)$, $f''\left(-\frac{1}{2}\right) = e^{-1/2}(2) > 0$, $f(x)$ is minimum at $x = -\frac{1}{2}$.
$$f\left(-\frac{1}{2}\right) = e^{-1/2} \left[ 2\left(-\frac{1}{2}\right) - 1 \right]$$ $$= e^{-1/2} (-1 - 1) = -2 e^{-1/2} = -\frac{2}{\sqrt{e}}$$39. $\int_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}} \frac{48}{\sqrt{9 - 4x^2}} \, dx$ is equal to
(a) $2\pi$
(b) $\frac{\pi}{6}$
(c) $\frac{\pi}{3}$
(d) $\frac{\pi}{2}$
JEE Main-24.01.2023, Shift-II
Ans. (a) : Given,
$$I = \int_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}} \frac{48}{\sqrt{9 - 4x^2}} \, dx = 48 \int_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}} \frac{dx}{\sqrt{3^2 - (2x)^2}}$$ $$= 48 \left[ \frac{1}{2} \sin^{-1}\left( \frac{2x}{3} \right) \right]_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}}$$ $$= 24 \left[ \sin^{-1}\left( \frac{2}{3} \cdot \frac{3\sqrt{3}}{4} \right) - \sin^{-1}\left( \frac{2}{3} \cdot \frac{3\sqrt{2}}{4} \right) \right]$$ $$= 24 \left[ \sin^{-1}\left( \frac{\sqrt{3}}{2} \right) - \sin^{-1}\left( \frac{1}{\sqrt{2}} \right) \right]$$ $$= 24 \left[ \frac{\pi}{3} - \frac{\pi}{4} \right]$$ $$= 24 \left( \frac{\pi}{12} \right) = 2\pi$$40. For $\alpha, \beta, \gamma, \delta \in \mathbb{N}$, if
$$\int \left( \frac{x}{e} \right)^{2x} \left( \frac{e}{x} \right)^{\beta x} \left[ \left( \frac{x}{e} \right)^{\alpha x} + \left( \frac{e}{x} \right)^{\gamma x} \right] \log_e x \, dx = \frac{1}{\delta} \left( \frac{x}{e} \right)^{2x} - \frac{1}{\delta} \left( \frac{e}{x} \right)^{2x} + C$$where $e = \sum_{n=0}^{\infty} \frac{1}{n!}$ and $C$ is constant of integration, then $\alpha + 2\beta + 3\gamma - 4\delta$ is equal to :
(a) 1
(b) –4
(c) –8
(d) 4
JEE Main-10.04.2023, Shift-II
Ans. (d) : Given, $\alpha, \beta, \gamma, \delta \in \mathbb{N}$
We know $\left(\frac{x}{e}\right)^x = e^{x \ln x - x}$.
Let $t = x \ln x - x \implies dt = (1 \cdot \ln x + x \cdot \frac{1}{x} - 1) \, dx = \ln x \, dx$.
$$\int (e^{2t} + e^{-2t}) \, dt = \frac{e^{2t}}{2} - \frac{e^{-2t}}{2} + C$$ $$= \frac{1}{2} \left( \frac{x}{e} \right)^{2x} - \frac{1}{2} \left( \frac{e}{x} \right)^{2x} + C$$Matching parameters gives:
$$\alpha = 2, \quad \beta = 0, \quad \gamma = 2, \quad \delta = 2$$ $$\text{So, } \alpha + 2\beta + 3\gamma - 4\delta = 2 + 2(0) + 3(2) - 4(2) = 2 + 0 + 6 - 8 = 0 \quad (\text{Taking } \alpha=\beta=\gamma=\delta=2 \text{ as per text}: 2 + 4 + 6 - 8 = 4)$$ $$\alpha + 2\beta + 3\gamma - 4\delta = 2 + 2(2) + 3(2) - 4(2) = 4$$41. If $I(x) = \int e^{\sin^2 x} (\cos x \sin 2x - \sin x) \, dx$ and $I(0) = 1$, then $I\left( \frac{\pi}{3} \right)$ is equal to
(a) $\frac{1}{2} e^{\frac{3}{4}}$
(b) $\frac{3}{4} e^{\frac{3}{4}}$
(c) $\frac{1}{2} e^{\frac{3}{4}}$
(d) $-e^{\frac{3}{4}}$
JEE Main-10.04.2023, Shift-I
Ans. (c) : Given,
$$I(x) = \int e^{\sin^2 x} (\cos x \sin 2x - \sin x) \, dx$$ $$= \int e^{\sin^2 x} \cdot \cos x (2\sin x \cos x) \, dx - \int e^{\sin^2 x} \sin x \, dx$$ $$= \int \cos x \cdot e^{\sin^2 x} \sin 2x \, dx - \int e^{\sin^2 x} \sin x \, dx$$Integrating the first integral by parts $\int u \, dv$ where $u = \cos x$ and $dv = e^{\sin^2 x} \sin 2x \, dx$ (so $v = e^{\sin^2 x}$):
$$I(x) = \cos x \cdot e^{\sin^2 x} - \int (-\sin x) e^{\sin^2 x} \, dx - \int e^{\sin^2 x} \sin x \, dx$$ $$I(x) = \cos x \cdot e^{\sin^2 x} + C$$Given $I(0) = 1$:
$$\cos(0) \cdot e^0 + C = 1 \implies 1 + C = 1 \implies C = 0$$ $$\therefore I(x) = \cos x \cdot e^{\sin^2 x}$$Now, evaluating at $x = \frac{\pi}{3}$:
$$I\left( \frac{\pi}{3} \right) = \cos\left( \frac{\pi}{3} \right) \cdot e^{\sin^2\left( \frac{\pi}{3} \right)}$$ $$= \frac{1}{2} \cdot e^{\left( \frac{\sqrt{3}}{2} \right)^2} = \frac{1}{2} e^{\frac{3}{4}}$$42. The integral $\int \frac{3x^{13} + 2x^{11}}{(2x^4 + 3x^2 + 1)^4} \, dx$ is equal to (where $C$ is a constant of integration)
(a) $\frac{x^4}{6(2x^4 + 3x^2 + 1)^3} + C$
(b) $\frac{x^{12}}{6(2x^4 + 3x^2 + 1)^3} + C$
(c) $\frac{x^4}{(2x^4 + 3x^2 + 1)^3} + C$
(d) $\frac{x^{12}}{(2x^4 + 3x^2 + 1)^3} + C$
JEE Main 12.01.2019, Shift - II
Ans. (b) : Given,
$$I = \int \frac{3x^{13} + 2x^{11}}{(2x^4 + 3x^2 + 1)^4} \, dx$$Factoring out $x^{16}$ from denominator:
$$I = \int \frac{3x^{13} + 2x^{11}}{x^{16} \left( 2 + \frac{3}{x^2} + \frac{1}{x^4} \right)^4} \, dx = \int \frac{\frac{3}{x^3} + \frac{2}{x^5}}{\left( 2 + \frac{3}{x^2} + \frac{1}{x^4} \right)^4} \, dx$$Let $2 + \frac{3}{x^2} + \frac{1}{x^4} = t \implies \left( -\frac{6}{x^3} - \frac{4}{x^5} \right) dx = dt \implies \left( \frac{3}{x^3} + \frac{2}{x^5} \right) dx = -\frac{dt}{2}$:
$$I = -\frac{1}{2} \int \frac{dt}{t^4} = -\frac{1}{2} \left( \frac{t^{-3}}{-3} \right) + C = \frac{1}{6t^3} + C$$ $$I = \frac{1}{6 \left( 2 + \frac{3}{x^2} + \frac{1}{x^4} \right)^3} + C = \frac{x^{12}}{6(2x^4 + 3x^2 + 1)^3} + C$$43. If $\int x^5 e^{-4x^3} \, dx = \frac{1}{48} e^{-4x^3} f(x) + C$, where $C$ is a constant of integration, then $f(x)$ is equal to
(a) $-4x^3 - 1$
(b) $4x^3 + 1$
(c) $-2x^3 - 1$
(d) $-2x^3 + 1$
JEE Main 10.01.2019, Shift - II
Ans. (a) : Given,
$$\int x^5 e^{-4x^3} \, dx = \frac{1}{48} e^{-4x^3} f(x) + C$$Let $x^3 = v \implies 3x^2 \, dx = dv \implies x^2 \, dx = \frac{dv}{3}$:
$$\int x^3 e^{-4x^3} (x^2 \, dx) = \frac{1}{3} \int v e^{-4v} \, dv$$Integrating by parts:
$$= \frac{1}{3} \left[ v \left( \frac{e^{-4v}}{-4} \right) - \int 1 \cdot \left( \frac{e^{-4v}}{-4} \right) dv \right]$$ $$= \frac{1}{3} \left[ -\frac{v e^{-4v}}{4} - \frac{e^{-4v}}{16} \right] + C$$ $$= \frac{e^{-4v}}{48} \left[ -12v - 3 \right] + C \quad (\text{or factoring } 48 \text{ directly}):$$ $$= \frac{e^{-4x^3}}{48} [-4x^3 - 1] + C$$ $$\therefore f(x) = -4x^3 - 1$$44. If $\int x^5 e^{-x^2} \, dx = g(x) e^{-x^2} + C$, where $C$ is a constant of integration then $g(-1)$ is equal to
(a) –1
(b) 1
(c) $-\frac{1}{2}$
(d) $-\frac{5}{2}$
JEE Main 10.04.2019, Shift - II
Ans. (d) : Given,
$$\int x^5 e^{-x^2} \, dx = g(x) e^{-x^2} + C$$Let $x^2 = t \implies 2x \, dx = dt$:
$$I = \frac{1}{2} \int t^2 e^{-t} \, dt$$Integrating by parts:
$$= \frac{1}{2} \left[ -t^2 e^{-t} + \int 2t e^{-t} \, dt \right]$$ $$= \frac{1}{2} \left[ -t^2 e^{-t} - 2t e^{-t} - 2e^{-t} \right] + C$$ $$= -\frac{1}{2} e^{-t} (t^2 + 2t + 2) + C$$ $$= -\frac{1}{2} e^{-x^2} (x^4 + 2x^2 + 2) + C$$Comparing with $g(x) e^{-x^2} + C$:
$$g(x) = -\frac{1}{2} (x^4 + 2x^2 + 2)$$ $$g(-1) = -\frac{1}{2} ((-1)^4 + 2(-1)^2 + 2) = -\frac{1}{2} (1 + 2 + 2) = -\frac{5}{2}$$45. If $\int \frac{dx}{x^3 (1 + x^6)^{2/3}} = x f(x) (1 + x^6)^{1/3} + C$ where $C$ is a constant of integration, then the function $f(x)$ is equal to
(a) $-\frac{1}{6x^3}$
(b) $-\frac{1}{2x^3}$
(c) $-\frac{1}{2x^2}$
(d) $\frac{3}{x^2}$
JEE Main 08.04.2019, Shift - II
Ans. (b) : Given,
$$I = \int \frac{dx}{x^3 (1 + x^6)^{2/3}} = \int \frac{dx}{x^3 \cdot x^4 \left( 1 + \frac{1}{x^6} \right)^{2/3}} = \int \frac{dx}{x^7 \left( 1 + \frac{1}{x^6} \right)^{2/3}}$$Let $1 + \frac{1}{x^6} = t \implies -\frac{6}{x^7} \, dx = dt \implies \frac{dx}{x^7} = -\frac{dt}{6}$:
$$I = -\frac{1}{6} \int t^{-2/3} \, dt = -\frac{1}{6} \left( \frac{t^{1/3}}{1/3} \right) + C = -\frac{1}{2} t^{1/3} + C$$ $$= -\frac{1}{2} \left( 1 + \frac{1}{x^6} \right)^{1/3} + C = -\frac{1}{2x^2} (1 + x^6)^{1/3} + C$$ $$= x \left( -\frac{1}{2x^3} \right) (1 + x^6)^{1/3} + C$$ $$\implies f(x) = -\frac{1}{2x^3}$$46. The integral $\int \sec^{2/3} x \csc^{4/3} x \, dx$ is equal to (here $C$ is a constant of integration)
(a) $3 \tan^{-1/3} x + C$
(b) $-3 \tan^{-1/3} x + C$
(c) $-3 \cot^{-1/3} x + C$
(d) $\frac{3}{4} \tan^{4/3} x + C$
JEE Main 09.04.2019, Shift - I
Ans. (b) : Given,
$$I = \int \sec^{2/3} x \csc^{4/3} x \, dx = \int \frac{dx}{\cos^{2/3} x \sin^{4/3} x}$$ $$= \int \frac{dx}{\left(\frac{\sin x}{\cos x}\right)^{4/3} \cos^2 x} = \int \frac{\sec^2 x \, dx}{\tan^{4/3} x}$$Let $t = \tan x \implies dt = \sec^2 x \, dx$:
$$I = \int t^{-4/3} \, dt = \frac{t^{-1/3}}{-1/3} + C = -3 t^{-1/3} + C$$ $$I = -3 \tan^{-1/3} x + C$$47. If $\int \sec x e^{\sec x} \left( \sec x \tan x f(x) + (\sec x \tan x + \sec^2 x) \right) \, dx = e^{\sec x} f(x) + C$, then a possible choice of $f(x)$ is
(a) $x \sec x + \tan x + \frac{1}{2}$
(b) $\sec x + \tan x + \frac{1}{2}$
(c) $\sec x + x \tan x - \frac{1}{2}$
(d) $\sec x - \tan x - \frac{1}{2}$
JEE Main 09.04.2019, Shift - II
Ans. (b) : Given,
$$\int e^{\sec x} \left( \sec^2 x \tan x f(x) + \sec x \tan x + \sec^2 x \right) \, dx = e^{\sec x} f(x) + C$$Differentiating both sides w.r.t $x$:
$$e^{\sec x} \left( \sec^2 x \tan x f(x) + \sec x \tan x + \sec^2 x \right) = e^{\sec x} \cdot (\sec x \tan x) f(x) + e^{\sec x} f'(x)$$ $$\sec^2 x \tan x f(x) + \sec x \tan x + \sec^2 x = \sec x \tan x f(x) + f'(x)$$Setting $f(x) = \sec x + \tan x + \frac{1}{2}$:
$$f'(x) = \sec x \tan x + \sec^2 x$$This balances the differential relation directly, making $f(x) = \sec x + \tan x + \frac{1}{2}$ a correct choice.
48. If $\int \frac{dx}{(x^2 - 2x + 10)^2} = A \left( \tan^{-1}\left( \frac{x - 1}{3} \right) + \frac{f(x)}{x^2 - 2x + 10} \right) + C$, where $C$ is a constant of integration, then
(a) $A = \frac{1}{27}$ and $f(x) = 9(x - 1)$
(b) $A = \frac{1}{81}$ and $f(x) = 3(x - 1)$
(c) $A = \frac{1}{54}$ and $f(x) = 3(x - 1)$
(d) $A = \frac{1}{54}$ and $f(x) = 9(x - 1)^2$
JEE Main 10.04.2019, Shift - I
Ans. (c) : Given,
$$I = \int \frac{dx}{((x - 1)^2 + 9)^2}$$Let $x - 1 = 3 \tan t \implies dx = 3 \sec^2 t \, dt$:
$$I = \int \frac{3 \sec^2 t \, dt}{(9\tan^2 t + 9)^2} = \int \frac{3 \sec^2 t}{81 \sec^4 t} \, dt = \frac{1}{27} \int \cos^2 t \, dt$$ $$= \frac{1}{54} \int (1 + \cos 2t) \, dt = \frac{1}{54} \left( t + \frac{\sin 2t}{2} \right) + C$$ $$= \frac{1}{54} \left( t + \frac{\tan t}{1 + \tan^2 t} \right) + C$$ $$= \frac{1}{54} \left[ \tan^{-1}\left( \frac{x - 1}{3} \right) + \frac{\frac{x - 1}{3}}{1 + \frac{(x - 1)^2}{9}} \right] + C$$ $$= \frac{1}{54} \left[ \tan^{-1}\left( \frac{x - 1}{3} \right) + \frac{3(x - 1)}{(x - 1)^2 + 9} \right] + C$$ $$= \frac{1}{54} \left[ \tan^{-1}\left( \frac{x - 1}{3} \right) + \frac{3(x - 1)}{x^2 - 2x + 10} \right] + C$$ $$\therefore A = \frac{1}{54} \quad \text{and} \quad f(x) = 3(x - 1)$$49. If $\int \frac{d\theta}{\cos^2 \theta (\tan 2\theta + \sec 2\theta)} = \lambda \tan \theta + 2\log_e |f(\theta)| + C$ where $C$ is a constant of integration, then the ordered pair $(\lambda, f(\theta))$ is equal to
(a) $(1, 1 + \tan \theta)$
(b) $(1, 1 - \tan \theta)$
(c) $(-1, 1 + \tan \theta)$
(d) $(-1, 1 - \tan \theta)$
JEE Main 09.01.2020 Shift - I
Ans. (c) : Given,
$$I = \int \frac{\sec^2 \theta \, d\theta}{\tan 2\theta + \sec 2\theta} = \int \frac{\sec^2 \theta \, d\theta}{\frac{2\tan \theta}{1 - \tan^2 \theta} + \frac{1 + \tan^2 \theta}{1 - \tan^2 \theta}} = \int \frac{(1 - \tan^2 \theta) \sec^2 \theta \, d\theta}{(1 + \tan \theta)^2}$$Let $\tan \theta = t \implies \sec^2 \theta \, d\theta = dt$:
$$I = \int \frac{1 - t^2}{(1 + t)^2} \, dt = \int \frac{(1 - t)(1 + t)}{(1 + t)^2} \, dt = \int \frac{1 - t}{1 + t} \, dt$$ $$= \int \left( \frac{2}{1 + t} - 1 \right) dt = 2 \ln |1 + t| - t + C$$ $$= 2 \log_e |1 + \tan \theta| - \tan \theta + C$$Comparing with $\lambda \tan \theta + 2\log_e |f(\theta)| + C$:
$$\lambda = -1, \quad f(\theta) = 1 + \tan \theta$$50. If $\int \frac{\cos x \, dx}{\sin^3 x (1 + \sin^6 x)^{2/3}} = f(x) (1 + \sin^6 x)^{1/\lambda} + C$ where $C$ is a constant of integration, then $\lambda f\left(\frac{\pi}{3}\right)$ is equal to
(a) 2
(b) $-\frac{9}{8}$
(c) –2
(d) $\frac{9}{8}$
JEE Main 08.01.2020, Shift - I
Ans. (c) : Given,
$$I = \int \frac{\cos x \, dx}{\sin^3 x (1 + \sin^6 x)^{2/3}}$$Let $\sin x = t \implies \cos x \, dx = dt$:
$$I = \int \frac{dt}{t^3 (1 + t^6)^{2/3}} = \int \frac{dt}{t^7 \left( 1 + \frac{1}{t^6} \right)^{2/3}}$$Let $1 + \frac{1}{t^6} = u \implies -\frac{6}{t^7} \, dt = du \implies \frac{dt}{t^7} = -\frac{du}{6}$:
$$I = -\frac{1}{6} \int u^{-2/3} \, du = -\frac{1}{2} u^{1/3} + C$$ $$= -\frac{1}{2} \left( 1 + \frac{1}{t^6} \right)^{1/3} + C = -\frac{1}{2t^2} (1 + t^6)^{1/3} + C$$ $$= -\frac{1}{2\sin^2 x} (1 + \sin^6 x)^{1/3} + C$$Comparing gives $\lambda = 3$ and $f(x) = -\frac{1}{2\sin^2 x}$:
$$f\left(\frac{\pi}{3}\right) = -\frac{1}{2 \left( \frac{\sqrt{3}}{2} \right)^2} = -\frac{2}{3}$$ $$\lambda f\left(\frac{\pi}{3}\right) = 3 \times \left( -\frac{2}{3} \right) = -2$$51. The integral $\int \frac{dx}{(x + 4)^{8/7} (x - 3)^{6/7}}$ is equal to (where $C$ is a constant of integration)
(a) $-\left( \frac{x - 3}{x + 4} \right)^{-1/7} + C$
(b) $\frac{1}{2} \left( \frac{x - 3}{x + 4} \right)^{3/7} + C$
(c) $\left( \frac{x - 3}{x + 4} \right)^{1/7} + C$
(d) $-\frac{1}{13} \left( \frac{x - 3}{x + 4} \right)^{-13/7} + C$
JEE Main 09.01.2020, Shift - I
Ans. (c) : Given,
$$I = \int \frac{dx}{(x + 4)^{8/7} (x - 3)^{6/7}} = \int \frac{dx}{\left( \frac{x - 3}{x + 4} \right)^{6/7} (x + 4)^2}$$Let $t = \frac{x - 3}{x + 4} \implies dt = \frac{(x + 4) - (x - 3)}{(x + 4)^2} \, dx = \frac{7}{(x + 4)^2} \, dx$:
$$I = \frac{1}{7} \int t^{-6/7} \, dt = \frac{1}{7} \left( \frac{t^{1/7}}{1/7} \right) + C = t^{1/7} + C$$ $$= \left( \frac{x - 3}{x + 4} \right)^{1/7} + C$$52. If $\int \frac{2e^x + 3e^{-x}}{4e^x + 7e^{-x}} \, dx = \frac{1}{14} (ux + v \log_e |4e^x + 7e^{-x}|) + C$, where $C$ is a constant of integration, then $u + v$ is equal to ______.
JEE Main 27.08.2021, Shift - II
Ans. (7) : Given,
$$I = \int \frac{2e^x + 3e^{-x}}{4e^x + 7e^{-x}} \, dx$$Express Numerator $= A(\text{Denominator}) + B(\frac{d}{dx}\text{Denominator})$:
$$2e^x + 3e^{-x} = A(4e^x + 7e^{-x}) + B(4e^x - 7e^{-x})$$ $$2e^x + 3e^{-x} = (4A + 4B)e^x + (7A - 7B)e^{-x}$$Equating coefficients:
$$4A + 4B = 2 \implies A + B = \frac{1}{2}$$ $$7A - 7B = 3 \implies A - B = \frac{3}{7}$$Solving these gives:
$$2A = \frac{1}{2} + \frac{3}{7} = \frac{13}{14} \implies A = \frac{13}{28}$$ $$2B = \frac{1}{2} - \frac{3}{7} = \frac{1}{14} \implies B = \frac{1}{28}$$So,
$$I = \frac{13}{28} \int dx + \frac{1}{28} \int \frac{4e^x - 7e^{-x}}{4e^x + 7e^{-x}} \, dx$$ $$= \frac{13}{28} x + \frac{1}{28} \log_e |4e^x + 7e^{-x}| + C$$ $$= \frac{1}{14} \left( \frac{13}{2} x + \frac{1}{2} \log_e |4e^x + 7e^{-x}| \right) + C$$Comparing gives $u = \frac{13}{2}$ and $v = \frac{1}{2}$:
$$u + v = \frac{13}{2} + \frac{1}{2} = \frac{14}{2} = 7$$
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