Show that $\int_0^1\frac{\ln(1+x)}{1+x^2}\mathrm dx=\frac{\pi}8\ln 2$ 
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.
 A: I suppose what you could do is write $$\log (1 + \tan y) = \frac{1}{2} \log 2 + \log \sin (y + \tfrac{\pi}{4} ) - \log \cos y,$$ then transform the second term with $v = \pi/4 - y$ to obtain $$\int_{y=0}^{\pi/4} \log \sin (y + \tfrac{\pi}{4}) \, dy = \int_{v=\pi/4}^0 \log \sin (\tfrac{\pi}{2} - v) \, (-dv) = \int_{v=0}^{\pi/4} \log \cos v \, dv.$$  Then this cancels with the integral of the third term, and you are left with $$\frac{\pi}{8} \log 2$$ as claimed.
A: Here's a different solution.
Let
$$f=\int_0^1 \frac{\log (x+1)}{x^2+1} \, dx$$
Replacing the $\log$ by the defining integral in the form
$$\log (x+1)=\int_0^1 \frac{x}{x y+1} \, dy$$
we find a double integral form of $f$
$$f = \int _0^1\int _0^1\frac{x}{\left(x^2+1\right) (x y+1)}dydx$$
Exchanging now the order of integration, i.e. carrying out the x-integration first, which is elementary, gives
$$\int_0^1 \frac{x}{\left(x^2+1\right) (x y+1)} \, dx=\frac{\pi  y}{4 y^2+4}+\frac{\log (4)}{4 y^2+4}-\frac{\log (y+1)}{y^2+1}$$
The integral over y now gives us the negative of the original integral f and two further terms which can be integrated to give
$$\int_0^1 \left(\frac{\pi  y}{4 y^2+4}+\frac{\log (4)}{4 y^2+4}\right) \, dy=\frac{1}{4} \pi  \log (2)$$
Hence we find that
$$f=\frac{1}{4} \pi  \log (2)-f$$
or
$$f=\frac{1}{8} \pi  \log (2)$$
QED.
A: Without using the given hint use the substitution $u = \frac{\pi}{4} -x$ to get $$I = \int_0^{\pi/4} \log (1 + \tan x) \, \mathrm{d}x = \int_0^{\pi/4} \log \left(\frac{2}{1 + \tan u}\right) \, \mathrm{d}u = \frac{\pi}{4}\ln 2 - I$$ since $1 + \tan (\pi/4 - u) = 1 + \frac{1-\tan u}{1 + \tan u} = \frac{2}{1 + \tan u}$.
So $2I = \frac{\pi}{4}\ln 2 \iff I = \frac{\pi}{8}\ln 2$
A: $\displaystyle J=\int_0^1 \dfrac{\ln(1+x)}{1+x^2}dx$
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,
$2J=\dfrac{\pi \ln 2}{4}$
$\boxed{J=\dfrac{\pi \ln 2}{8}}$
A: Ok, I solved it, sorry :p. 
We can observe that $\sin(x+\pi/4)$ is symmetric to $\cos x$ in $[0,\pi/4]$, so we can write
$$\begin{align}\int_0^{\pi/4}\ln(\sqrt 2\sin(x+\pi/4)/\cos x)\mathrm dx&=\int_0^{\pi/4}\frac12\ln 2\mathrm dx+\int_{\pi/4}^{\pi/2}\ln\sin x\mathrm dx-\int_0^{\pi/4}\ln\cos x\mathrm dx\\&=\frac{\ln2}2\int_0^{\pi/4}\mathrm dx=\frac{\pi}8\ln 2\end{align}$$
