The value of the integral _0^ dx (1+ x ^2 )(1+ x )^2 is - Vidyadip

MCQ

The value of the integral $\int_0^{\infty} \frac{ dx }{\left(1+ x ^2\right)(1+ x )^2}$ is

A
$\frac{1}{4}$
✓
$\frac{1}{2}$
C
$\frac{3}{4}$
D
$\infty$

✓

Answer

Correct option: B.

$\frac{1}{2}$

b
(b)

Put $x =\tan \theta$

$dx =\sec ^2 \theta d \theta$

$I=\int \limits_0^{\pi / 2} \frac{\sec ^2 \theta}{\left(1+\tan ^2 \theta\right)(1+\tan \theta)^2} d \theta$

$=\int \limits_0^{\pi / 2} \frac{\cos ^2 \theta}{(\sin \theta+\cos \theta)^2} d \theta$

$I+I=\int \limits_0^{\pi / 2} \frac{d \theta}{(\sin \theta+\cos \theta)^2}$

$=\int \limits_0^{\pi / 2} \frac{\sec ^2 \theta}{(\tan \theta+1)^2} d \theta$

$=-\left.\frac{1}{(1+\tan \theta)}\right|_0 ^{\frac{\pi}{2}}=1$

$\Rightarrow I =\frac{1}{2}$

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