TITLE: Find the following integral (most likely substitution)
QUESTION [6 upvotes]: $$\int_0^1 \frac{\ln(1+x^2)}{1+x^2} \ dx$$
I tried letting $x^2=\tan \theta$ but it didn't work. What should I do?
Please don't give full solution, just a hint and I will continue.

REPLY [2 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{1}{\ln\pars{1 +x^{2}} \over 1 + x^{2}}\,\dd x:\ {\large ?}}$

With $\ds{x \equiv \tan\pars{\theta}}$:
  \begin{align}&\color{#c00000}{\int_{0}^{1}{\ln\pars{1 +x^{2}} \over 1 + x^{2}}
\,\dd x}
=-\int_{0}^{\pi/4}\ln\pars{\cos^{2}\pars{\theta}}\,\dd\theta
\\[3mm]&=-\int_{0}^{\pi/4}\ln\pars{{2\sin\pars{\theta}\cos\pars{\theta} \over 2}\,{\cos\pars{\theta} \over \sin\pars{\theta}}}\,\dd\theta
=-\int_{0}^{\pi/4}\ln\pars{\half\,\sin\pars{2\theta}\cot\pars{\theta}}\,\dd\theta
\\[3mm]&=-\int_{0}^{\pi/4}\ln\pars{\cot\pars{\theta}}\,\dd\theta
-\half\int_{0}^{\pi/2}\ln\pars{\half\,\sin\pars{\theta}}\,\dd\theta
\\[3mm]&=-G + {\pi \over 4}\,\ln\pars{2}
-\half\,\color{#00f}{\int_{0}^{\pi/2}\ln\pars{\sin\pars{\theta}}\,\dd\theta}
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\pars{1}
\end{align}
  where we used
  one of the Integral Representations of Catalan Constant $\ds{G}$. Namely
  $$
G \equiv \int_{0}^{\pi/4}\ln\pars{\cot\pars{\theta}}\,\dd\theta
$$

Let's evaluate the $\ds{\color{#00f}{\mbox{remaining integral}}}$:
\begin{align}&\color{#00f}{\int_{0}^{\pi/2}\ln\pars{\sin\pars{\theta}}
\,\dd\theta}
=\half\bracks{\int_{0}^{\pi/2}\ln\pars{\sin\pars{\theta}}\,\dd\theta
+\int_{0}^{\pi/2}\ln\pars{\sin\pars{{\pi \over 2} - \theta}}\,\dd\theta}
\\[3mm]&=\half\int_{0}^{\pi/2}\ln\pars{\half\,\sin\pars{2\theta}}\,\dd\theta
={1 \over 4}\int_{0}^{\pi}\ln\pars{\half\,\sin\pars{\theta}}\,\dd\theta
\\[3mm]&={1 \over 4}\int_{0}^{\pi/2}\ln\pars{\half\,\sin\pars{\theta}}
\,\dd\theta
+{1 \over 4}\int_{\pi/2}^{\pi}\ln\pars{\half\,\sin\pars{\theta}}\,\dd\theta
\\[3mm]&=\half\int_{0}^{\pi/2}\ln\pars{\half\,\sin\pars{\theta}}\,\dd\theta
=-\,{\pi \over 4}\,\ln\pars{2}
+\half\color{#00f}{\int_{0}^{\pi/2}\ln\pars{\sin\pars{\theta}}\,\dd\theta}
\\[3mm]&\imp\quad
\color{#00f}{\int_{0}^{\pi/2}\ln\pars{\sin\pars{\theta}}\,\dd\theta}
=-\,{\pi \over 2}\,\ln\pars{2}
\end{align}

This result is replaced in $\pars{1}$:
  $$\color{#66f}{\large\int_{0}^{1}{\ln\pars{1 +x^{2}} \over 1 + x^{2}}
\,\dd x = \half\,\pi\ln\pars{2} - G}
$$