how to solve $\int _0^1\frac{\ln \left(1+x\right)}{a^2+x^2}\:\mathrm{d}x$ 
how to solve $$\mathcal{J(a)}=\int _0^1\frac{\ln \left(1+x\right)}{a^2+x^2}\:\mathrm{d}x$$

i used the differentiation under the integral and got
\begin{align}
\mathcal{J(b)}&=\int _0^1\frac{\ln \left(1+bx\right)}{a^2+x^2}\:\mathrm{d}x
\\[3mm]
\mathcal{J'(b)}&=\int _0^1\frac{x}{\left(a^2+x^2\right)\left(1+bx\right)}\:\mathrm{d}x
\\[3mm]
&=\frac{a^2b}{1+a^2b^2}\int _0^1\frac{1}{a^2+x^2}\:\mathrm{d}x+\frac{1}{1+a^2b^2}\int _0^1\frac{x}{a^2+x^2}\:\mathrm{d}x-\frac{b}{1+a^2b^2}\int _0^1\frac{1}{1+bx}\:\mathrm{d}x
\\[3mm]
&=\frac{ab}{1+a^2b^2}\operatorname{atan} \left(\frac{1}{a}\right)+\frac{1}{2}\frac{\ln \left(1+a^2\right)}{1+a^2b^2}-\frac{\ln \left(a\right)}{1+a^2b^2}-\frac{\ln \left(1+b\right)}{1+a^2b^2}
\end{align}
But we know that $\mathcal{J}(1)=\mathcal{J(a)}$ and $\mathcal{J}(0)=0$
\begin{align}
\int_0^1\mathcal{J'(b)}\:\mathrm{d}b&=a\:\operatorname{atan} \left(\frac{1}{a}\right)\int _0^1\frac{b}{1+a^2b^2}\:\mathrm{d}b+\frac{\ln \left(1+a^2\right)}{2}\int _0^1\frac{1}{1+a^2b^2}\:\mathrm{d}b-\ln \left(a\right)\int _0^1\frac{1}{1+a^2b^2}\:\mathrm{d}b
\\
&-\int _0^1\frac{\ln \left(1+b\right)}{1+a^2b^2}\:\mathrm{d}b
\\[3mm]
\mathcal{J(a)}&=\frac{1}{2a}\operatorname{atan} \left(\frac{1}{a}\right)\ln \left(1+a^2\right)+\frac{1}{2a}\ln \left(1+a^2\right)\operatorname{atan} \left(a\right)-\frac{1}{a}\ln \left(a\right)\:\operatorname{atan} \left(a\right)-\underbrace{\int _0^1\frac{\ln \left(1+b\right)}{1+a^2b^2}\:\mathrm{d}b}_{\mathcal{I}}
\end{align}
but how to calculate ${\mathcal{I}}$, i tried using the same technique but it didnt work
 A: Repeating your calculations using Feynman's trick
$$J'(b)=\frac{-\log \left(a^2\right)+\log \left(a^2+1\right)+2 a b \cot ^{-1}(a)-2 \log
   (b+1)}{2 (a^2 b^2+1)}$$
$$J(b)=\int\frac{-\log \left(a^2\right)+\log \left(a^2+1\right)+2 a b \cot ^{-1}(a)}{2 (a^2 b^2+1)}\,db-\int\frac{\log(b+1)}{ a^2 b^2+1}\,db$$ The first integral is simple
$$J_1(b)=\int\frac{-\log \left(a^2\right)+\log \left(a^2+1\right)+2 a b \cot ^{-1}(a)}{2 (a^2 b^2+1)}\,db$$
$$J_1(b)=\frac{\cot ^{-1}(a) \log \left(a^2 b^2+1\right)+\left(\log
   \left(a^2+1\right)-\log \left(a^2\right)\right) \tan ^{-1}(a b)}{2 a}$$
$$K_1=\int_0 ^1\frac{-\log \left(a^2\right)+\log \left(a^2+1\right)+2 a b \cot ^{-1}(a)}{2 (a^2 b^2+1)}\,db$$
$$K_1=\frac{\left(\log \left(a^2+1\right)-\log \left(a^2\right)\right) \tan
   ^{-1}(a)+\log \left(a^2+1\right) \cot ^{-1}(a)}{2 a}$$
More tedious is the second integral
$$J_2(b)=\int\frac{\log(b+1)}{ a^2 b^2+1}\,db$$ $$J_2(b)=\frac{i \left(\text{Li}_2\left(\frac{a
   (b+1)}{a-i}\right)-\text{Li}_2\left(\frac{a (b+1)}{a+i}\right)+\log (b+1)
   \left(\log \left(1-\frac{a (b+1)}{a-i}\right)-\log \left(1-\frac{a
   (b+1)}{a+i}\right)\right)\right)}{2 a}$$
$$K_2(b)=\int_0^1\frac{\log(b+1)}{ a^2 b^2+1}\,db=$$
$$\frac{i \left(-\text{Li}_2\left(\frac{a}{a-i}\right)+\text{Li}_2\left(\frac{2
   a}{a-i}\right)+\text{Li}_2\left(\frac{a}{a+i}\right)-\text{Li}_2\left(\frac{2
   a}{a+i}\right)+\log (2) \left(\log \left(-\frac{a+i}{a-i}\right)-\log
   \left(-\frac{a-i}{a+i}\right)\right)\right)}{2 a}$$ which, for sure, is a real number.
Remark
I think that this could have benn done without Feynman's trick
$$\frac{\log \left(1+x\right)}{a^2+x^2}=\frac{\log \left(1+x\right)}{(x+i a)(x-i a)}$$
$$\int \frac{\log \left(1+x\right)}{a^2+x^2}\,dx=\frac i {2a}\left(\int \frac{\log \left(1+x\right)}{x+i a}\,dx-\int \frac{\log \left(1+x\right)}{x-i a}\,dx  \right)$$
$$\int \frac{\log \left(1+x\right)}{x+i a}\,dx=\text{Li}_2\left(\frac{x+1}{1-i a}\right)+\log (x+1) \log \left(1-\frac{x+1}{1-ia}\right)$$
$$\int \frac{\log \left(1+x\right)}{x-i a}\,dx=\text{Li}_2\left(\frac{x+1}{1+i a}\right)+\log (x+1) \log \left(1-\frac{x+1}{1+ia}\right)$$
A: The integral $I$ has been evaluated here by Felix Marin.
Using it yields the closed form of the integral
\begin{align}
J\left(a\right)&=\int_0^1\frac{\ln\left(1+x\right)}{a^2+x^2}\:dx=\frac{1}{2a}\operatorname{arctan} \left(\frac{1}{a}\right)\ln \left(1+a^2\right)+\frac{1}{2a}\ln \left(1+a^2\right)\operatorname{arctan} \left(a\right)
\\
&-\frac{1}{a}\ln \left(a\right)\:\operatorname{arctan} \left(a\right)-2\ln \left(2\right)\frac{\operatorname{arctan} \left(a\right)}{a}-\frac{1}{a}\mathfrak{I}\left(\operatorname{Li}_2\left(\frac{2a}{i+a}\right)-\operatorname{Li}_2\left(\frac{a}{i+a}\right)\right)
\end{align}
Complex analysis seems like the only way to go with this integral because software also gives closed forms featuring the imaginary unit with polylogs, i also checked if this worked for particular values and it does, though i'd still like to see if anyone can come up with an approach without any complex methods.
A: 
Suppose $\left(a,b,c,z\right)\in\mathbb{R}\times\mathbb{R}_{>0}\times\mathbb{R}_{\ge0}\times\mathbb{R}_{>0}$, and set
$$\alpha:=\arctan{\left(\frac{a}{b}\right)}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right),$$
$$\gamma:=\arctan{\left(\frac{c-a}{b}\right)}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right),$$
$$\theta:=\arctan{\left(\frac{z+a}{b}\right)}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$$
Then, the following logarithmic integral can be evaluated in terms of Clausen functions using the general result derived here:
$$\begin{align}
\int_{0}^{z}\mathrm{d}x\,\frac{2b\ln{\left(x+c\right)}}{\left(x+a\right)^{2}+b^{2}}
&=\operatorname{Cl}_{2}{\left(2\alpha+2\gamma\right)}-\operatorname{Cl}_{2}{\left(2\theta+2\gamma\right)}+\operatorname{Cl}_{2}{\left(\pi-2\alpha\right)}-\operatorname{Cl}_{2}{\left(\pi-2\theta\right)}\\
&~~~~~+\left(\theta-\alpha\right)\ln{\left(b^{2}\sec^{2}{\left(\gamma\right)}\right)},\\
\end{align}$$
where the Clausen function (of order 2) is defined for real arguments by the integral representation
$$\operatorname{Cl}_{2}{\left(\vartheta\right)}:=-\int_{0}^{\vartheta}\mathrm{d}\varphi\,\ln{\left(\left|2\sin{\left(\frac{\varphi}{2}\right)}\right|\right)};~~~\small{\vartheta\in\mathbb{R}}.$$

Applying the formula to the particular case where $a=0\land c=1\land z=1$, and with a little help from the duplication formula for the Clausen function, we obtain the following delightfully compact result:
$$\begin{align}
\forall b\in\mathbb{R}_{>0}:\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(x+1\right)}}{x^{2}+b^{2}}
&=\frac{4\theta\ln{\left(\csc{\left(\theta\right)}\right)}-\operatorname{Cl}_{2}{\left(4\theta\right)}}{4b};~~~\small{\theta:=\arctan{\left(\frac{1}{b}\right)}\in\left(0,\frac{\pi}{2}\right)}.\\
\end{align}$$
It should be clear that Clausen functions are the natural tool to use if you're trying to stick to real methods.

