Help me evaluate $\int_0^1 \frac{\log(x+1)}{1+x^2} dx$ I need to evaluate this integral: $\int_0^1 \frac{\log(x+1)}{1+x^2} dx$.
I've tried $t=\log(x+1)$, $t=x+1$, but to no avail. I've noticed that:
$\int_0^1 \frac{\log(x+1)}{1+x^2} dx = \int_0^1\log(x+1) \arctan'(x)dx =\left. \log(x+1)\arctan(x) \right|_{x=0}^{x=1} - \int_0^1\frac{\arctan(x)}{x+1}dx$
But can't get further than this.
Any help is appreciated, thank you.
 A: Let 
$$I(a)=\int_0^1 \frac{\log(1+ax)}{1+x^2}dx$$
Differentiate it, to get
$$I'(a)=\int_0^1 \frac{x}{(1+ax)(1+x^2)}dx$$
Integrate that rational function, then integrate w.r.t. $a$ and find $I(a=1)$.
As Theorem suggested, you can also do the following:
$$\int_0^1 \frac{\log(1+x)}{1+x^2}dx$$
$$\int_0^1 \left(\int_0^x \frac{1}{1+y}dy\right)\frac{1}{1+x^2}dx$$
$$\int_0^1 \int_0^x \frac{1}{1+y}dy\frac{1}{1+x^2}dx$$
Now make a change of variables $y=ux$ in the inner integral:
$$\tag 1 \int_0^1 \int_0^1 \frac{x}{1+ux}\frac{1}{1+x^2}dudx$$
Now partial fractions:
$${x \over {1 + xu}}{1 \over {1 + {x^2}}} = {1 \over {1 + {u^2}}}{x \over {1 + {x^2}}} + {u \over {1 + {u^2}}}{1 \over {1 + {x^2}}} - {u \over {1 + {u^2}}}{1 \over {1 + xu}}$$
Now, integrating the first two terms, which account to the same$^1$, gives that you integral is
$$\mathcal I=\frac \pi 4\log 2-\int_0^1\int_0^1 \frac u {1+u^2}\frac{1}{1+xu}dxdu$$
Now, the latter integral is just our original integral, due to symmetry, as you see in $(1)$ 
This means that $$\mathcal I =\frac \pi 8 \log 2$$
as desired.
$1$: symmetry, once again.

See here for a similar integral and its solution with double integrals.
Some insight about the two methods considered:
Note that as Sasha is showing
$$I(1)=\int_0^1 I'(a)da=\int_0^1 \int_0^1\frac{x}{(1+ax)(1+x^2)}dxda$$ which is exaclty what we got in the second option
$$I=\int_0^1\int_0^1 \frac{1   }{1+mx}\frac{x}{1+x^2}dxdm$$
This means any way you find to solve any of the two will indeed solve the other.
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
\color{#00f}{\large\int_{0}^{1}{\ln\pars{1 + x} \over 1 + x^{2}}\,\dd x}&
=\int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{\theta}}\,\dd\theta
\\[3mm]&=\half\bracks{%
\int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{\theta}}\,\dd\theta
+
\int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{{\pi \over 4} - \theta}}\,\dd\theta}
\\[3mm]&=\half\bracks{%
\int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{\theta}}\,\dd\theta
+
\int_{0}^{\pi/4}\ln\pars{%
1 + {1 - \tan\pars{\theta} \over 1 + \tan\pars{\theta}}}\,\dd\theta}
\\[3mm]&=\half\bracks{%
\int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{\theta}}\,\dd\theta
+
\int_{0}^{\pi/4}\ln\pars{2\over 1 + \tan\pars{\theta}}\,\dd\theta}
\\[3mm]&=\half\int_{0}^{\pi/4}\ln\pars{2}\,\dd\theta
=\color{#00f}{\large{1 \over 8}\,\pi\ln\pars{2}}
\end{align}
A: $\displaystyle A=\int_0^1\dfrac{\Big(\log(1+x)\Big)^2}{1+x^2}dx$
Perform the change of variable $y=\dfrac{1-x}{1+x}$
$\displaystyle A=\int_0^1\dfrac{\left(\log\left(\dfrac{2}{1+x}\right)\right)^2}{1+x^2}dx=\int_0^1\dfrac{\Big(\log 2-\log(1+x)\Big)^2}{1+x^2}dx$
$\displaystyle A=\int_0^1\dfrac{\Big(\log 2\Big)^2}{1+x^2}dx-2\int_0^1\dfrac{\log 2\log(1+x)}{1+x^2}dx+A$
Therefore,
$\displaystyle \int_0^1\dfrac{2\log 2\log(1+x)}{1+x^2}dx=\int_0^1\dfrac{\Big(\log 2\Big)^2}{1+x^2}dx$
Finally,
$\displaystyle \int_0^1\dfrac{\log(1+x)}{1+x^2}dx=\int_0^1\dfrac{\log 2}{2(1+x^2)}dx=\dfrac{\log 2}{2}\Big[\arctan x\Big]_0^1=\dfrac{\pi\log 2}{8}$
A: For these integrals are very useful substitutions homograph type.
To note $\displaystyle I(a) = \int^{a}_{0}\frac{\ln(x+a)}{x^{2}+a^2}dx.$
Using the substitution $x=\dfrac{-at+a^2}{t+a} = u(t)$ with $u'(t)=-\dfrac{2a^2}{(t+a)^2} $ we find 
$$\begin{align*}I(a) &= \int^{0}_{a}\frac{\ln\left(\frac{-at+a^2}{t+a}+a\right)}{\left(\frac{-at+a^2}{t+a}\right)^2+a^2}\left(- \frac{2a^2}{(t+a)^2}\right)dt=\\
&= \int^{a}_{0}\frac{\ln 2a^2 - \ln(t+a)}{t^2+a^2}dt = \\
&=\int^{a}_{0}\frac{\ln 2a^2}{t^2+a^2}dt -\int^{a}_{0}\frac{\ln(t+a)}{t^2+a^2}dt=\\ &
=\frac{\ln2a^{2}}{a} \arctan 1- I(a).\end{align*}$$  
And finally $$I(a) = \dfrac{\pi}{8}\cdot\frac{\ln2a^{2}}{a}.$$
For $ a= 1$ we obtain $I(1) = \dfrac{\pi}{8}\ln2.$
See also: http://www.recreatiimatematice.ro/arhiva/articole/RM12011DICU.pdf
A: Going a little round-about way. Consider, for $ s \geqslant 0$, a parametric modification of the integral at hand: 
$$
  \mathcal{I}(s) = \int_0^1 \frac{\log(1+s x)}{1+x^2} \mathrm{d} x
$$
The goal is to determine $\mathcal{I}(1)$. Now:
$$ \begin{eqnarray}
   \mathcal{I}(1) &=& \int_0^1 \mathcal{I}^\prime(s) \mathrm{d} s = \int_0^1 \left( \int_0^1 \frac{x}{1+s x} \frac{\mathrm{d} x}{1+x^2} \right) \mathrm{d} s \\
 &=& \int_0^1 \left.\left( - \frac{1}{1+s^2} \log(1+s x) + \frac{s}{1+s^2} \arctan(x)   + \frac{1}{2} \frac{\log(1+x^2)}{1+x^2}  \right) \right|_{x=0}^{x=1} \mathrm{d} s \\
   &=& \int_0^1 \left( \color\green{ -\frac{\log(1+s)}{1+s^2}} + \frac{1}{4} \frac{\pi s+\log(4)}{1+s^2}\right) \mathrm{d} s = - \mathcal{I}(1) + \frac{1}{4} \pi \log(2)
\end{eqnarray}
$$
Hence
$$
   \mathcal{I}(1) = \frac{\pi}{8} \log(2)
$$
A: Here's a solution that uses simpler tools (or at least tools that I'm more familiar with):
$I =\int_0^1\frac{\log(1+x)}{1+x^2} dx $. 
We change $x$ into $x=\tan(t)$. Then $t=\arctan{x}$, $dt=\frac{1}{1+x^2}dx$, and the integral becomes:
$I = \int_0^{\frac{\pi}{4}}\log(1+\tan(t))dt$. Now $s = \frac{\pi}{4}-t$, $ds=-dt$, and the integral becomes:
$I = -\int_{\frac{\pi}{4}}^0 \log(1+\tan(\frac{\pi}{4}-s))ds = \int_0^{\frac{\pi}{4}} \log(1+\tan(\frac{\pi}{4}-s))ds$ 
Using $\tan(a-b) = \frac{\tan a - \tan b}{1 + \tan(a)\tan(b)}$, we have
$I = \int_0^{\frac{\pi}{4}} \log(1+\frac{1 - \tan s}{1 + \tan s}) ds = \int_0^{\frac{\pi}{4}}\log(\frac{2}{1+\tan s}) ds = \int_0^{\frac{\pi}{4}} (\log(2) - \log(1+\tan s)) ds = \int_0^{\frac{\pi}{4}}\log(2)ds - I = \frac{\pi}{4}\log(2) - I$.
So $I = \frac{\pi}{4}\log(2) - I$, hence $I = \frac{\pi}{8}\log(2)$.
