The value of _0^1 (1+x) 1+x^2 d x is - Vidyadip

MCQ

The value of $\int_0^1 \frac{\log (1+x)}{1+x^2} d x$ is

A
$\frac{\pi}{2} \log 2$
B
$\pi \log 2$
C
$\log 2$
✓
$\frac{\pi}{8} \log 2$

✓

Answer

Correct option: D.

$\frac{\pi}{8} \log 2$

(D)
Let $I =\int_0^1 \frac{\log (1+x)}{1+x^2} d x$
Put $x=\tan \theta \Rightarrow d x=\sec ^2 \theta d \theta$
$\therefore \quad I=\int_0^{\pi / 4} \frac{\log (1+\tan \theta)}{1+\tan ^2 \theta} \cdot \sec ^2 \theta d \theta$
$=\int_0^{\pi / 4} \frac{\log (1+\tan \theta)}{\sec ^2 \theta} \cdot \sec ^2 \theta d \theta$
$\therefore \quad I=\int_0^{\pi / 4} \log (1+\tan \theta) d \theta$
$\int_0^{\pi / 4} \log (1+\tan x) d x=\frac{\pi}{8} \log 2$
$\Rightarrow I=\frac{\pi}{8} \log 2 $

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