TITLE: Show that $\int_0^1\frac{\ln(1+x)}{1+x^2}\mathrm dx=\frac{\pi}8\ln 2$
QUESTION [8 upvotes]: Show that $\int_0^1\frac{\ln(1+x)}{1+x^2}\mathrm dx=\frac{\pi}8\ln 2$ using the change of variable $x=\tan y$.
Hint: $1+\tan x=\sqrt 2\sin(x+\pi/4)/\cos x$.

With the change of variable suggested I get
$$\int_0^1\frac{\ln(1+x)}{1+x^2}\mathrm dx=\int_0^{\pi/4}\ln(1+\tan y)\mathrm dy$$
but I dont know exactly what to do here or what to do with the identity $1+\tan x=\sqrt 2\sin(x+\pi/4)/\cos x$.
I tried some changes of variable or integration by parts but nothing work. I tried to write the logarithm as a series but the elements $\tan^k x$ are complicate to integrate. Some help will be appreciated, thank you.

REPLY [4 votes]: $\displaystyle J=\int_0^1 \dfrac{\ln(1+x)}{1+x^2}dx$
Perform the change of variable $y=\dfrac{1-x}{1+x}$,
$\begin{align}\displaystyle J&=\int_0^1 \dfrac{\ln\left(\tfrac{2}{1+x}\right)}{1+x^2}dx\\
&=\int_0^1 \dfrac{\ln 2}{1+x^2}dx-J\\
&=\dfrac{\pi \ln 2}{4}-J
\end{align}$
Therefore,
$2J=\dfrac{\pi \ln 2}{4}$
$\boxed{J=\dfrac{\pi \ln 2}{8}}$