_1^2 dx/((x+1)(x+2)) = Option 1: _e 8 - _e 9 Option 2: _e 9 - _e 8 Option 3: _e 8 + _e 9 Option 4: 2_e 9 - _e 8

Answer: _e 9 - _e 8. Solution: Partial fractions give 1/(x+1)-1/(x+2), so the integral is [, (x+1)/(x+2), ]_1^2=34-23=98=_e9-_e8.

Calculus

Integrals

Medium

common

$\int_1^2 \frac{dx}{(x+1)(x+2)} =$

$\log_e 8 - \log_e 9$

$\log_e 9 - \log_e 8$

$\log_e 8 + \log_e 9$

$2\log_e 9 - \log_e 8$

✅ Correct Option: 2

Partial fractions give $\frac{1}{x+1}-\frac{1}{x+2}$ , so the integral is $\left[\log\left|\frac{x+1}{x+2}\right|\right]_1^2=\log\frac34-\log\frac23=\log\frac98=\log_e9-\log_e8$ .

2025: 29 May Shift 2

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

2026: 18 May Shift 1

Match List-I with List-II

List-I	List-II
(A) $\displaystyle\int_{\pi/6}^{\pi/3} \cos x \, dx$	(I) $0$
(B) $\displaystyle\int_{-\pi}^{\pi} (\sin^{61}x + x^{123}) \, dx$	(II) $4$
(C) $\displaystyle\int_{0}^{\pi/2} \dfrac{1}{1+\tan^3 x} \, dx$	(III) $\dfrac{\sqrt{3}-1}{2}$
(D) $\displaystyle\int_{-2}^{2} (2 - \left\vert x \right\vert) \, dx$	(IV) $\dfrac{\pi}{4}$

Choose the correct answer from the options given below:

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: 26 May Shift 2

$\int_{1}^{2} \frac{1}{x(x+1)} dx, x > 0$ equals

2025: 30 May Shift 1

$\int \frac{1}{(x + 1)(x + 2)} dx$ is equal to

2022: 7 Aug Shift 2

$\int \frac{x^2 + 4}{(x^2 + 3)(x^2 + 5)} \, dx = u \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + v \tan^{-1}\left(\frac{x}{\sqrt{5}}\right) + c$ , where c is arbitrary constant, then value of $\frac{1}{u^2} + \frac{1}{v^2}$ is equal to :

∫12dx(x+1)(x+2)=\int_1^2 \frac{dx}{(x+1)(x+2)} =∫12​(x+1)(x+2)dx​=

Solution

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$\int_1^2 \frac{dx}{(x+1)(x+2)} =$