Question

Prove that: ${\int }_0^1\frac{\log (1+x)}{(1+x^2)}dx=\frac{\pi }{8}\log 2$.

Answer 1

Let $I$ be the integral given to us.

$\therefore I={\int }_0^1\frac{\log 1+x}{1+x^2}dx$

Substitute $x=\tan t⟹dx={\sec }^{2}tdt$

$\therefore I={\int }_0^{\frac{\pi }{4}}\frac{\log (1+\tan t)}{{\sec }^{2}t}{\sec }^{2}tdt$

$\therefore I={\int }_0^{\frac{\pi }{4}}\log (1+\tan t)dt$

Now, as ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$, we have

$I={\int }_0^{\frac{\pi }{4}}\log (1+\tan (\frac{\pi }{4}-t))dt$

$I={\int }_0^{\frac{\pi }{4}}\log (1+\frac{1-\tan t}{1+\tan t})dt$ (using $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$)

$\therefore I={\int }_0^{\frac{\pi }{4}}\log \left(\frac{2}{1+\tan t}\right)dt$

$\therefore I={\int }_0^{\frac{\pi }{4}}\left[\log \right(2)-\log (1+\tan t\left)\right]dt$

$\therefore I={\int }_0^{\frac{\pi }{4}}\log \left(2\right)dt-I$

$\therefore 2I={\int }_0^{\frac{\pi }{4}}\log \left(2\right)dt$

$\therefore 2I=t\log \left(2\right){|}_0^{\frac{\pi }{4}}$

$\therefore 2I=\frac{\pi }{4}\log \left(2\right)$

$\therefore I=\frac{\pi }{8}\log \left(2\right)$

Hence, proved.