_ - 1 ^1 x^3 + |x| + 3 x^2 + 4|x| + 3 dx is equal to - - Vidyadip

MCQ

$\int\limits_{ - 1}^1 {\frac{{{x^3} + |x| + 3}}{{{x^2} + 4|x| + 3}}dx} $ is equal to -

A
$\frac{4}{\pi }\int\limits_0^{\frac{\pi }{2}} {\log (\sin \alpha )d\alpha } $
B
$ - \frac{4}{\pi }\int\limits_0^{\frac{\pi }{2}} {\log (\sin \theta )d\alpha } $
C
$ - \frac{2}{\pi }\int\limits_0^{\frac{\pi }{2}} {\log (\sin 2\alpha )d\alpha } $
✓
$ - \frac{2}{\pi }\int\limits_0^{\frac{\pi }{2}} {\log (\sin \alpha ) + \log (\cos \alpha )d\alpha } $

✓

Answer

Correct option: D.

$ - \frac{2}{\pi }\int\limits_0^{\frac{\pi }{2}} {\log (\sin \alpha ) + \log (\cos \alpha )d\alpha } $

d
$=\int_{-1}^{1} \frac{x^{3}}{(|x|+1)(x |+3)}+\int_{-1}^{1} \frac{|x|+3}{(|x|+1)(|x|+3)} d x$

$=0+2 \int_{0}^{1} \frac{1}{(x)+1} d x=2 \ln 2$

Use $\int_{0}^{\pi / 2} \log (\sin \alpha) d \alpha=\int_{0}^{\pi / 2} \log (\cos \alpha) d \alpha=-\frac{\pi}{2} \ln 2$

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