Evaluate: _0^1 x ^ -1 x (1+x^2 )^ 3 / 2 d x - Vidyadip

MCQ

Evaluate: $\int_0^1 \frac{x \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}} d x$

A
$\frac{4-\pi}{2 \sqrt{2}}$
B
$\frac{4+\pi}{2 \sqrt{2}}$
✓
$\frac{4-\pi}{4 \sqrt{2}}$
D
None of these

✓

Answer

Correct option: C.

$\frac{4-\pi}{4 \sqrt{2}}$

(c): Let $\int_0^1 \frac{x \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}} d x$ Put $\tan ^{-1} x=\theta \Rightarrow x=\tan \theta \Rightarrow d x=\sec ^2 \theta d \theta$ When, $x=0 \Rightarrow \theta=0$ and $x=1 \Rightarrow \theta=\frac{\pi}{4}$ $I=\int_0^1 \frac{x \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}} d x=\int_0^{\pi / 4} \frac{\theta \tan \theta}{\sec ^3 \theta} \sec ^2 \theta d \theta$$=\int_0^{\pi / 4} \theta \sin \theta d \theta=[-\theta \cos \theta]_0^{\pi / 4}-\int_0^{\pi / 4}(-\cos \theta) d \theta$[Integrating by parts]
$=[-\theta \cos \theta]_0^{\pi / 4}+[\sin \theta]_0^{\pi / 4}=\frac{4-\pi}{4 \sqrt{2}}$

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