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

Answer: 1/2((x^2+3)/(x^2+4)) + C. Solution: Let u = x^2, du = 2x\ dx. Then x/((x^2+3)(x^2+4))dx = 1/2 du/((u+3)(u+4)). Using partial fractions: 1/((u+3)(u+4)) = 1/(u+3) - 1/(u+4). = 1/2, (u+3)/(u+4), + C = 1/2((x^2+3)/(x^2+4)) + C.

Calculus

Integrals

Medium

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

$\frac{1}{2}\log\left(\frac{x^2+4}{x^2+3}\right) + C$

$\log\left(\frac{x^2+3}{x^2+4}\right) + C$

$\frac{1}{2}\log\left(\frac{x^2+3}{x^2+4}\right) + C$

$\log\left(\frac{x^2+4}{x^2+3}\right) + C$

✅ Correct Option: 3

Let $u = x^2$ , $du = 2x\ dx$ . Then $\int \frac{x}{(x^2+3)(x^2+4)}dx = \frac{1}{2}\int \frac{du}{(u+3)(u+4)}$ .

Using partial fractions: $\frac{1}{(u+3)(u+4)} = \frac{1}{u+3} - \frac{1}{u+4}$ .

$= \frac{1}{2}\log\left|\frac{u+3}{u+4}\right| + C = \frac{1}{2}\log\left(\frac{x^2+3}{x^2+4}\right) + C$ .

2026: 15 May Shift 1

$\displaystyle\int \dfrac{dx}{(x^2+4)^{\frac{3}{2}}}$ is equal to ( where $c$ is an arbitrary constant)

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The value of $\int_{-1}^{1} \dfrac{x^6}{6^x+1}\,dx$ is

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The value of $\int_{-2}^{3} |x-1|\, dx$ is

2025: 13 May Shift 2

$\int \left(\frac{\cos x - \sin x}{1 + \sin 2x}\right) dx$ is equal to

2025: 30 May Shift 1

Match List-I with List-II

List-I	List-II
(A) $\int_0^1 \frac{x^2}{1 + x^3} dx$	(I) 0
(B) $\int_0^\pi 3\sin x dx$	(II) $2\log_e\left(\frac{3}{2}\right)$
(C) $\int_{-1}^1 \sin^5 x \cos^6 x dx$	(III) 6
(D) $\int_2^3 \frac{4}{x^2 - 1} dx$	(IV) $\frac{1}{3}\log_e 2$

Choose the correct answer from the options given below:

2025: 22 May Shift 1

The integral $\int e^x\left(\frac{x-1}{2x^2}\right)dx$ is equal to

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

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

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