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

Answer: 1/2_e C(x-1)(x-3). Solution: (x^2 - 4)/((x+2)(x-1)(x-3)) = ((x-2)(x+2))/((x+2)(x-1)(x-3)) = (x-2)/((x-1)(x-3)). Partial fractions: (x-2)/((x-1)(x-3)) = (1/2)/(x-1) + (1/2)/(x-3). Integral = 1/2, x-1, + 1/2, x-3, + C = 1/2_e, C(x-1)(x-3), .

Calculus

Integrals

Medium

$\int \frac{x^2 - 4}{(x+2)(x-1)(x-3)} dx =$

$2 \log |C(x-1)(x-3)|$

$\log_e |C(x-1)(x-3)|$

$\frac{1}{2}\log_e |C(x-1)(x-3)|$

$\frac{1}{2}\log_e |C(x+2)(x-3)|$

✅ Correct Option: 3

$\frac{x^2 - 4}{(x+2)(x-1)(x-3)} = \frac{(x-2)(x+2)}{(x+2)(x-1)(x-3)} = \frac{x-2}{(x-1)(x-3)}$ . Partial fractions: $\frac{x-2}{(x-1)(x-3)} = \frac{1/2}{x-1} + \frac{1/2}{x-3}$ . Integral $= \frac{1}{2}\log|x-1| + \frac{1}{2}\log|x-3| + C = \frac{1}{2}\log_e |C(x-1)(x-3)|$ .

2026: 15 May Shift 1

If C is an arbitrary constant, then

Match the LIST-I with LIST-II

	LIST-I		LIST-II
A.	$\displaystyle\int \sqrt{x^2+a^2}\, dx$	I.	$\dfrac{x}{2}\sqrt{a^2-x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} + C$
B.	$\displaystyle\int \sqrt{a^2-x^2}\, dx$	II.	$\log\left\vert x + \sqrt{x^2+a^2}\right\vert + C$
C.	$\displaystyle\int \dfrac{1}{\sqrt{x^2+a^2}}\, dx$	III.	$\sin^{-1}\dfrac{x}{a} + C$
D.	$\displaystyle\int \dfrac{1}{\sqrt{a^2-x^2}}\, dx$	IV.	$\dfrac{x}{2}\sqrt{x^2+a^2} + \dfrac{a^2}{2}\log\left\vert x + \sqrt{x^2+a^2}\right\vert + C$

Choose the correct answer from the options given below:

2022: 30 Aug Shift 1

$\int_{\frac{1}{3}}^{1} \frac{(x - x^3)^{\frac{1}{3}}}{x^4} dx =$

2026: 22 May Shift 1

Match List-I with List-II

Where C is arbitrary constant.

List-I	List-II
(A) $\displaystyle\int \dfrac{dx}{x^2 - a^2} =$	(I) $\log\left\vert x + \sqrt{x^2+a^2}\right\vert + C$
(B) $\displaystyle\int \dfrac{dx}{a^2 - x^2} =$	(II) $\log\left\vert x + \sqrt{x^2-a^2}\right\vert + C$
(C) $\displaystyle\int \dfrac{dx}{\sqrt{x^2 - a^2}} =$	(III) $\dfrac{1}{2a}\log\left\vert \dfrac{x-a}{x+a}\right\vert + C$
(D) $\displaystyle\int \dfrac{dx}{\sqrt{x^2 + a^2}} =$	(IV) $\dfrac{1}{2a}\log\left\vert \dfrac{a+x}{a-x}\right\vert + C$

Choose the correct answer from the options given below:

2026: 26 May Shift 2

$\displaystyle\int \sqrt[3]{x}(x^3+4x^2+7)\,dx$ is equal to-

2026: 29 May Shift 1

Match List-I with List-II

List-I	List-II
(A) $\displaystyle\int_{-1}^{1}(2x^5+3x^3+4x+1)dx=$	(I) 2
(B) $\displaystyle\int_{-2}^{2}(4x^3+2)dx=$	(II) 12
(C) $\displaystyle\int_{-1}^{1}x^3dx=$	(III) 8
(D) $\displaystyle\int_{-3}^{3}(3x^5+2)dx=$	(IV) 0

Choose the correct answer from the options given below:

2022: 23 Aug Shift 1

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

∫x2−4(x+2)(x−1)(x−3)dx=\int \frac{x^2 - 4}{(x+2)(x-1)(x-3)} dx =∫(x+2)(x−1)(x−3)x2−4​dx=

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

$\int \frac{x^2 - 4}{(x+2)(x-1)(x-3)} dx =$