Finding $ \int_0^1 \frac {\ln x}{1+x^2}\mathrm dx $ Today I encountered the problem of how to find $$ \displaystyle\int_{0}^{1} \frac {\ln x}{1 + x^2}\mathrm dx $$ but got no start on it. Is this one of those integrals which we have to approach from the definite aspect only? Wolfram Alpha suggests that the indefinite version is miserable. 
Thank you! 
 A: Use integration by parts to get $$ \displaystyle\int_0^1 \frac {\ln x}{1 + x^2} \, \mathrm{d}x = \left[ \arctan(x) \ln(x) \right]_0^1 - \displaystyle\int_0^1 \frac {\arctan(x)}{x} \, \mathrm{d}x = - \displaystyle\int_0^1 \frac {\arctan(x)}{x} \, \mathrm{d}x. $$Now, we use the Taylor expansion of $\arctan(x)$ to get $$ -\int_0^1 \sum_{n=0}^{\infty}\frac{1}{x}\frac{(-1)^nx^{2n+1}}{2n+1}\,\mathrm{d}x $$Switch the sum and the integral and it should be easy to end at the result $ -\displaystyle\sum_{n=0}^{\infty} \frac {(-1)^n}{(2n+1)^2} $. 
This is now the Catalan constant. 
A: let $-I = \int_0^1{\ln x \over 1 + x^2} \ dx.$  we will need $$a_k = \int_0^\infty xe^{-kx}\ dx = {1 \over k^2} \int_0^\infty xe^{-x} \ dx = {1 \over k^2}$$
$I =  \int_0^1{\ln x \over 1 + x^2} \ dx,$ by a change of variable
$ u = -\ln x, x = e^{-u}, $ the integral 
$\begin{eqnarray} I &=& \int_0^\infty{ ue^{-u} \over 1 + e^{-2u}} \ du \\
&=& \int_0^\infty ue^{-u}\left(1 - e^{-2u} + e^{-4u}+ \cdots \right) du\\
&=& a_1 - a_3 + a_5 + \cdots \\
&=&1 - {1 \over 3^2} + {1 \over 5^2} -{1 \over 7^2}+ \cdots = 0.9159\cdots=
\mbox{ Catalan constant}
\end{eqnarray}$
p.s.: i just learned the name for this constant.
A: \begin{align}
\int_{0}^{1}\frac{\ln x}{1+x^2}\mathrm dx&=\int_0^{1}\sum_{n=1}^\infty(-1)^n\, x^{2n}\ln x\;\mathrm dx\tag1\\
&=\sum_{n=1}^\infty(-1)^n\int_0^{1} x^{2n}\ln x\;\mathrm dx\tag2\\
&=-\sum_{n=1}^\infty\frac{(-1)^n}{(2n+1)^2}\tag3\\
&=\bbox[5pt,border:3px #FF69B4 solid]{\color{red}{\large-\text{G}}}
\end{align}

Explanation :
$(1)\;$ Use series expansion $\;\displaystyle \frac{1}{1+y}=\sum_{n=1}^\infty(-1)^n \,y^{n}$ and replace $y=x^2$
$(2)\;$ Use formula $\;\displaystyle \int_0^1 x^n \ln^k x\ dx=\frac{(-1)^k\, k!}{(n+1)^{k+1}}$, for $\  k=0,1,2,\cdots$
$(3)\;$ Series expression of Catalan's constant
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large\int_{0}^{1}{\ln\pars{x} \over 1 + x^{2}}\,\dd x}
=\Re\int_{0}^{1}{\ln\pars{x} \over 1 - \ic x}\,\dd x
=\Im\int_{0}^{1}{\ln\pars{-\ic\bracks{\ic x}} \over 1 - \ic x}
\,\pars{\ic\,\dd x}
=\Im\int_{0}^{\ic}{\ln\pars{-\ic x} \over 1 - x}\,\dd x
\\[5mm]&=\Im\bracks{\left.\vphantom{\LARGE A}
-\ln\pars{1 - x}\ln\pars{-\ic x}\right\vert_{\, 0}^{\,\ic}
+\int_{0}^{\ic}{\ln\pars{1 - x} \over x}\,\dd x}
=-\,\Im\int_{0}^{\ic}\Li{2}'\pars{x}\,\dd x
=-\,\Im\Li{2}\pars{\ic}
\\[5mm]&=-\,\Im\sum_{n\ =\ 1}^{\infty}{\ic^{n} \over n^{2}}
=-\,\Im\sum_{n\ =\ 1}^{\infty}{\ic^{2n - 1} \over \pars{2n - 1}^{2}}
=-\,\Im\bracks{%
\ic^{-1}\sum_{n\ =\ 1}^{\infty}{\pars{-1}^{n} \over \pars{2n - 1}^{2}}}
=-\sum_{n\ =\ 0}^{\infty}{\pars{-1}^{n} \over \pars{2n + 1}^{2}}
\\[5mm]&=\color{#66f}{\Large-G}
\end{align}

where $\ds{G}$ is the
  Catalan Constant.

