$\displaystyle \int_0^{+\infty} \dfrac{\ln(x+1)}{x^2+1} \ \mathrm dx$ Find the value of:
$$\int_0^{+\infty} \dfrac{\ln(x+1)}{x^2+1} \ \mathrm dx$$
I tried using substitution but it doesn't work. Is there any way to solve it ? 
 A: Hint. One may set
$$
I(t):=\int_0^{+\infty} \dfrac{\ln(1+tx)}{x^2+1} \ \mathrm dx,\qquad t>0.
$$ Then one may obtain
$$
\begin{align}
I'(t)=&\int_0^{+\infty} \dfrac{x}{(x^2+1)(1+tx)} \ \mathrm dx
\\\\&=\int_0^{+\infty} \left(\dfrac{x}{(x^2+1)(1+tx)}-\frac{t}{\left(t^2+1\right) (t x+1)} \right) \mathrm dx
\\\\&=\left[\frac{\ln \left(x^2+1\right)}{2 t^2+2}+\frac{t \arctan x-\ln (t x+1)}{t^2+1} \right]_0^{+\infty} 
\\\\&=\frac{\pi t}{2 t^2+2}-\frac{2 \ln t}{2 t^2+2},\qquad t>0,
\end{align}
$$ giving, with $I(0)=0$,

$$
\int_0^{+\infty} \dfrac{\ln(1+x)}{x^2+1}=I(1)=I(1)-I(0)=\int_0^1I'(t)\:dt=C+\frac{\pi}{4} \ln 2
$$

where $C$ is the Catalan constant.
A: 
I tried using substitution but it doesn't work.

A trig substitution will do. Letting $x = \tan \theta$, the integral becomes $\int_0^{\pi/2} \ln(\tan \theta + 1)\, d\theta$, which is the same as 
$$\int_0^{\pi/2} \ln(\sin \theta + \cos \theta)\, d\theta - \int_0^{\pi/2}\ln(\cos \theta)\, d\theta$$
Since $\sin(\theta + \cos \theta) = \sqrt{2}\sin(\theta + \pi/4)$,
\begin{align}\int_0^{\pi/2}\ln(\sin \theta + \cos \theta)\, d\theta &= \int_0^{\pi/2} \ln\sqrt{2}\, d\theta + \int_0^{\pi/2} \ln\sin(\theta + \pi/4)\, d\theta\\
&= \frac{\pi}{4}\ln2 + \int_0^{\pi/2}\ln\sin(\theta + \pi/4)\, d\theta
\end{align}
Note $\int_0^{\pi/4}\ln \sin(\theta + \pi/4)\, d\theta = \int_{\pi/4}^{\pi/2} \ln \sin\theta\,d\theta$ and $$\int_{\pi/4}^{\pi/2} \ln\sin(\theta + \pi/4)\, d\theta = \int_{\pi/4}^{\pi/2}\ln \cos(\theta - \pi/4)\, d\theta = \int_0^{\pi/4}\ln\cos \theta\, d\theta$$
The total integral is then
$$\frac{\pi}{4}\ln 2 + \int_{\pi/4}^{\pi/2} \ln \sin \theta\, d\theta + \int_{0}^{\pi/4} \ln\cos \theta\, d\theta - \int_0^{\pi/2}\ln \cos \theta\, d\theta$$
or
$$\frac{\pi}{4}\ln 2 + \int_{\pi/4}^{\pi/2}\ln \tan\theta\, d\theta$$
The latter integral equals $\int_0^{\pi/4}\tan(\pi/2 - \theta)\, d\theta = \int_0^{\pi/4}\ln \cot \theta = C$, Catalan's constant. Hence, the integral evaluates to 

$$\frac{\pi}{4}\ln 2 + C$$

A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{0}^{\infty}{\ln\pars{x + 1} \over x^{2} + 1}\,\dd x & =
\int_{0}^{1}{\ln\pars{1 + x} \over 1 + x^{2}}\,\dd x +
\bracks{\int_{1}^{\infty}{\ln\pars{1/x + 1} \over x^{2} + 1}\,\dd x + \int_{1}^{\infty}{\ln\pars{x} \over x^{2} + 1}\,\dd x}
\end{align}

Note that the last integral is the Catalan Constant $C$. Namely,
  $\ds{\int_{1}^{\infty}{\ln\pars{x} \over x^{2} + 1}\,\dd x = C}$.

Lets $\ds{x\ \mapsto\ 1/x}$ in the second integral:
\begin{align}
\int_{0}^{\infty}{\ln\pars{x + 1} \over x^{2} + 1}\,\dd x - C & =
\int_{0}^{1}{\ln\pars{1 + x} \over 1 + x^{2}}\,\dd x +
\int_{1}^{0}{\ln\pars{x + 1} \over 1/x^{2} + 1}\pars{-\,{1 \over x^{2}}}\,\dd x
\\[5mm] & =
2\int_{0}^{1}{\ln\pars{1 + x} \over 1 + x^{2}}\,\dd x
\,\,\,\stackrel{x\ =\ \tan\pars{\theta}}{=}\,\,\,
2\int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{\theta}}\,\dd\theta
\\[5mm] & =
\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
\\[5mm] & =
\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
\\[5mm] & =
\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
\\[5mm] & =
\int_{0}^{\pi/4}\ln\pars{2}\,\dd\theta = {\pi \over 4}\,\ln\pars{2}
\\[5mm] \implies &\
\bbox[15px,#ffe,border:1px dotted navy]{\ds{%
\int_{0}^{\infty}{\ln\pars{x + 1} \over x^{2} + 1}\,\dd x =
{\pi \over 4}\,\ln\pars{2} + C}}
\end{align}
