e^x (1/x - 1/x^2) dx = Option 1: e^x * 1/x + C Option 2: -e^x * 1/x + C Option 3: e^x + C Option 4: e^x * 1/x^2 + C

Answer: e^x * 1/x + C. Solution: Using e^x[f(x) + f'(x)]dx = e^x f(x) + C. Here f(x) = 1/x, f'(x) = -1/x^2. So integral = e^x * 1/x + C.

Calculus

Integrals

Easy

$\int e^x \left(\frac{1}{x} - \frac{1}{x^2}\right) dx =$

$e^x \cdot \frac{1}{x} + C$

$-e^x \cdot \frac{1}{x} + C$

$e^x + C$

$e^x \cdot \frac{1}{x^2} + C$

✅ Correct Option: 1

Using $\int e^x[f(x) + f'(x)]dx = e^x f(x) + C$ . Here $f(x) = \frac{1}{x}$ , $f'(x) = -\frac{1}{x^2}$ . So integral $= e^x \cdot \frac{1}{x} + C$ .

2026: 19 May Shift 1

Match List-I with List-II

[c is an arbitrary constant]

List-I	List-II
(A) $\int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x}\,dx$	(I) $\tan x - x + c$
(B) $\int \frac{\sec^2 x}{\text{cosec}^2 x}\,dx$	(II) $2\tan x - 3\sec x + c$
(C) $\int \sec x(\sec x + \tan x)\,dx$	(III) $\tan x + \cot x + c$
(D) $\int \frac{2-3\sin x}{\cos^2 x}\,dx$	(IV) $\tan x + \sec x + c$

Choose the correct answer from the options given below:

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Match List-I with List-II

List-I	List-II
Indefinite integral	Solution (where 'c' is an arbitrary constant)
(A) $\displaystyle\int \sqrt{x^2-a^2} \, dx$	(I) $\dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + c$
(B) $\displaystyle\int \dfrac{1}{x^2-a^2} \, dx$	(II) $\dfrac{x}{2}\sqrt{x^2+a^2} + \dfrac{a^2}{2}\log\left[x+\sqrt{x^2+a^2}\right] + c$
(C) $\displaystyle\int \dfrac{1}{x^2+a^2} \, dx$	(III) $\dfrac{x}{2}\sqrt{x^2-a^2} - \dfrac{a^2}{2}\log\left[x+\sqrt{x^2-a^2}\right] + c$
(D) $\displaystyle\int \sqrt{x^2+a^2} \, dx$	(IV) $\dfrac{1}{2a}\log\left\vert \dfrac{x-a}{x+a} \right\vert + C$

Choose the correct answer from the options given below:

2025: 16 May Shift 1

Match List-I with List-II

List-I	List-II
Functions	Integrals
---	---
(A) $\int \frac{dx}{x^2 - 4}, x \neq \pm 2$	(I) $\log \vert x + \sqrt{4 + x^2}\vert + C$ , where $C$ is an arbitrary constant
(B) $\int \frac{1}{\sqrt{16 - x^2}} dx; \vert x\vert < 4$	(II) $\sin^{-1} \left( \frac{x}{4} \right) + C$ , where $C$ is an arbitrary constant
(C) $\int \frac{1}{16 + x^2} dx$	(III) $\frac{1}{4} \log \left\vert \frac{x-2}{x+2} \right\vert + C$ , where $C$ is an arbitrary constant
(D) $\int \frac{1}{\sqrt{4 + x^2}} dx$	(IV) $\frac{1}{4} \tan^{-1} \left( \frac{x}{4} \right) + C$ , where $C$ is an arbitrary constant

Choose the correct answer from the options given below:

2025: 22 May Shift 1

$\int \frac{dx}{2\sin^2 x + 5\cos^2 x}$ is equal to

2025: 27 May Shift 1

$\int e^{-x}(\cot x + \cosec^2 x)dx =$

2024: 16 May Shift 1

$\int \mathrm{e}^{\mathrm{x}}\left(\frac{2 \mathrm{x}+1}{2 \sqrt{\mathrm{x}}}\right) \mathrm{dx}$