Question

The value of ${\int }_0^1\frac{{\log }_e(x+1)}{1+x^2}dx=\frac{\pi \ln a}{b}$, then $a^2+b^2$ equal to

• A. $20$
• B. $68$
• C. $40$
• D. $Noneofthese$

Answer 1

Answer: (B)

$68$

$\begin{array}{c}I={\int }_0^1\frac{{\log }_e\left(x+1\right)}{1+x^2}dx\\ x=\tan \theta \\ \frac{dx}{d\theta }=1+x^2\\ I={\int }_0^{\frac{\pi }{4}}{\log }_e\left(1+\tan \theta \right)d\theta \\ I={\int }_0^{\frac{\pi }{4}}{\log }_e\left(1+\tan \left(\frac{\pi }{4}-\theta \right)\right)d\theta \\ I={\int }_0^{\frac{\pi }{4}}{\log }_e\left(\frac{2}{1+\tan \theta }\right)d\theta \\ I={\int }_0^{\frac{\pi }{4}}{\log }_{e}2d\theta -{\int }_0^{\frac{\pi }{4}}{\log }_e\left(1+\tan \theta \right)d\theta \\ 2I={\log }_{e}2\frac{\pi }{4}\\ I=\frac{\pi }{4}\ln 2\\ a=2\\ and,b=8\\ Now,\\ a^2+b^2=4+64=68.\end{array}$

Hence, Option $B$ is the correct answer.