((x-3)e^x)/((x-1)^3)dx is equal to: (where C is an arbitrary constant) Option 1: 1/(e^x(x-1)^2)+C Option 2: ((x-1)^2)/e^x+C Option 3: (x-1)^2e^x+C Option 4: e^x/((x-1)^2)+C

Answer: e^x/((x-1)^2)+C. Solution: Write the integrand as e^x[f(x)+f'(x)] with f(x) = 1/((x-1)^2), since f'(x) = (-2)/((x-1)^3) and 1/((x-1)^2)-2/((x-1)^3) = (x-3)/((x-1)^3). Hence the integral is e^x/((x-1)^2)+C.

Calculus

Integrals

Medium

core

$\displaystyle\int \dfrac{(x-3)e^x}{(x-1)^3}dx$ is equal to:

(where C is an arbitrary constant)

$\dfrac{1}{e^x(x-1)^2}+C$

$\dfrac{(x-1)^2}{e^x}+C$

$(x-1)^2e^x+C$

$\dfrac{e^x}{(x-1)^2}+C$

✅ Correct Option: 4

Write the integrand as $e^x\left[f(x)+f'(x)\right]$ with $f(x) = \dfrac{1}{(x-1)^2}$ , since $f'(x) = \dfrac{-2}{(x-1)^3}$ and $\dfrac{1}{(x-1)^2}-\dfrac{2}{(x-1)^3} = \dfrac{x-3}{(x-1)^3}$ .

Hence the integral is $\dfrac{e^x}{(x-1)^2}+C$ .

2026: 31 May Shift 2

$\displaystyle\int \dfrac{1+\cos\theta}{1-\cos\theta}\,d\theta =$

2022: 23 Aug Shift 1

$\int \sqrt{x^2 - 4x + 5} \, dx =$

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

2025: 3 June Shift 2

If $\int \sqrt\frac{1-x}{{1+x}} dx = a\sqrt{1-x^2} + \beta \sin^{-1}x + C$ , Where C is an arbitrary constant, then which of the following are TRUE?

(A) $\alpha = 1$

(B) $\alpha = -1$

(D) $\beta = -1$

Choose the correct answer from the options given below:

2025: 15 May Shift 2

$\int \frac{(x^4 - x)^{1/4}}{x^5} dx$ is equal to

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:

∫(x−3)ex(x−1)3dx\displaystyle\int \dfrac{(x-3)e^x}{(x-1)^3}dx∫(x−1)3(x−3)ex​dx is equal to: (where C is an arbitrary constant)

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CUET Mathematics Past Year Questions (Topic-Wise)

$\displaystyle\int \dfrac{(x-3)e^x}{(x-1)^3}dx$ is equal to:

(where C is an arbitrary constant)