((x-1)e^x)/x^2 dx, x > 0 equals (where C is an arbitrary constant) Option 1: xe^x + C Option 2: e^x/x + C Option 3: (e^(-x))/x^2 + C Option 4: e^x + 1/x + C

Answer: e^x/x + C. Solution: Given: ((x-1)e^x)/x^2 dx, x > 0 Consider e^x/x and differentiate using the Quotient Rule: d/dx(e^x/x) = (x * d/dx(e^x) - e^x * d/dx(x))/x^2 = (x * e^x - e^x * 1)/x^2 = (xe^x - e^x)/x^2 = (e^x(x-1))/x^2 = ((x-1)e^x)/x^2 This matches the original integrand. Since d/dx(e^x/x) = ((x-1)e^x)/x^2 ((x-1)e^x)/x^2 dx = e^x/x + C

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$\int \frac{(x-1)e^x}{x^2} dx, x > 0$ equals (where C is an arbitrary constant)

$xe^x + C$

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

$\frac{e^{-x}}{x^2} + C$

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

✅ Correct Option: 2

Given: $\int \frac{(x-1)e^x}{x^2} dx, x > 0$

Consider $\frac{e^x}{x}$ and differentiate using the Quotient Rule:

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

$= \frac{x \cdot e^x - e^x \cdot 1}{x^2}$

$= \frac{xe^x - e^x}{x^2}$

$= \frac{e^x(x-1)}{x^2}$

$= \frac{(x-1)e^x}{x^2}$

This matches the original integrand.

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

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

2025: 3 June Shift 1

$\int \sqrt{1 + \frac{x^2}{9}} dx$ is equal to

(Where C is an arbitrary constant)

2026: 19 May Shift 2

$\int \frac{\log x - 3}{(\log x)^4} dx$ is equal to

(where C is an arbitrary constant and $\log_e x = \log x$ )

2025: 14 May Shift 2

If $\int \frac{dx}{(x-1)^3/^4. (x+2)^5/^4} = a[1 - g(x)]^b + c$ , where $c$ is a constant of integration, then which of the following are true?

(A) $a = \frac{2}{3}$

(B) $\beta = \frac{3}{4}$

(D) $g(x) = \frac{3}{(x+2)}$

Choose the correct answer from the options given below:

2026: 21 May Shift 1

Value of $\int (e^{2\log x} + e^{x\log 2} + e^{2\log 2})dx$ is

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: 2 June Shift 1

$\int_0^1 \frac{dx}{\sqrt{1+x} - \sqrt{x}}$ is equal to

∫(x−1)exx2dx,x>0\int \frac{(x-1)e^x}{x^2} dx, x > 0∫x2(x−1)ex​dx,x>0 equals (where C is an arbitrary constant)

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$\int \frac{(x-1)e^x}{x^2} dx, x > 0$ equals (where C is an arbitrary constant)