Question

The value of ${\int }_0^1\frac{8\log (1+x)}{1+x^2}dx$ is

• A. $2\pi \log 2$
• B. $\pi \log 2$
• C. $\log 2$
• D. None of these

Answer 1

The correct option is B $\pi \log 2$

$\begin{array}{c}Let,\\ I={\int }_0^1\frac{8\log \left(1+x\right)}{1+x^2}dx\\ \Rightarrow I=8{\int }_0^1\frac{\log \left(1+x\right)}{1+x^2}dx\end{array}$

$\begin{array}{c}putx=\tan \theta \\ \Rightarrow \frac{dx}{d\theta }={\sin }^2\theta \\ \Rightarrow dx={\sin }^2\theta d\theta \end{array}$

Also,

$x=0\Rightarrow \tan \theta =0\Rightarrow \theta =0$

and,

$x=1\Rightarrow \tan \theta =1\Rightarrow \theta =\frac{\pi }{4}$

Now

$\begin{array}{c}\therefore I=8{\int }_0^{\frac{\pi }{4}}\frac{\log \left(1+\tan \theta \right)}{1+{\tan }^2\theta }{\sin }^2\theta d\theta \\ =8{\int }_0^{\frac{\pi }{4}}\frac{\log \left(1+\tan \theta \right)}{{\sin }^2\theta }{\sin }^2\theta d\theta \\ =8{\int }_0^{\frac{\pi }{4}}\log \left(1+\tan \theta \right)d\theta \\ =8\times \frac{\pi }{8}\log 2\\ =\pi \log 2\end{array}$

Hence

$\therefore {\int }_0^1\frac{8\log \left(1+x\right)}{1+x^2}dx=\pi \log 2$