How to integrate $\int{\frac{\ln(1+t)}{t^2+1}}dt$ I was given this problem,
$$\int{\frac{x^2+1}{x^4-x^2+1}}\ln{(1+x-\frac{1}{x})}dx$$
Putting $x-\frac{1}{x}=t$,
We get $$\int{\frac{\ln(1+t)}{1+t^2}}dt$$
But I am struggling to integrate after this step? How should I continue?
Any hints would be helpful. Thank you.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\int{\ln\pars{1 + t} \over 1 + t^{2}}\,\dd t} =
\int{\ln\pars{1 + t} \over \pars{t + \ic}\pars{t - \ic}}\,\dd t \\[5mm] = &\
{1 \over 2}\int\ln\pars{1 + t}\pars{{1 \over t - \ic} + {1 \over t + \ic}}\,\dd t
\\[5mm] = &\
{1 \over 2}\sum_{\sigma = \pm 1}\int{\ln\pars{1 + t} \over t - \sigma\ic}\,\dd t
\,\,\,\stackrel{x\ \equiv\ t + 1}{=}\,\,\,
-\,{1 \over 2}\sum_{\sigma = \pm 1}\int{\ln\pars{x} \over
1 + \sigma\ic - x}\,\dd x
\\[5mm] \stackrel{y\ \equiv\ x/\pars{1 + \sigma\ic}}{=}\,\,\,&
-\,{1 \over 2}\sum_{\sigma = \pm 1}\int{\ln\pars{\bracks{1 + \sigma\ic}y} \over
1 - y}\,\dd y
\\[5mm] = &\
-\,{1 \over 2}\sum_{\sigma = \pm 1}\bracks{%
-\ln\pars{1 - y}\ln\pars{\bracks{1 + \sigma\ic}y} + \int{\ln\pars{1 - y} \over y}\,\dd y}
\\[5mm] = &\
{1 \over 2}\sum_{\sigma = \pm 1}\bracks{%
\ln\pars{1 - y}\ln\pars{\bracks{1 + \sigma\ic}y} + \mrm{Li}_{2}\pars{y}}
\\[5mm] = &\
{1 \over 2}\sum_{\sigma = \pm 1}\bracks{%
\ln\pars{1 - {x \over 1 + \sigma\ic}}\ln\pars{x} +
\mrm{Li}_{2}\pars{x \over 1 + \sigma\ic}}
\\[5mm] = &\
{1 \over 2}\sum_{\sigma = \pm 1}\bracks{%
\ln\pars{1 - {t + 1 \over 1 + \sigma\ic}}\ln\pars{t + 1} +
\mrm{Li}_{2}\pars{t + 1 \over 1 + \sigma\ic}}
\\[5mm] = &\
\bbx{{1 \over 2}\,\ln\pars{t + 1}\sum_{\sigma = \pm 1}
\ln\pars{1 - {t + 1 \over 1 + \sigma\ic}} +
{1 \over 2}\sum_{\sigma = \pm 1}\mrm{Li}_{2}\pars{t + 1 \over 1 + \sigma\ic}}
\end{align}
A: In order to find the value of $\int_{\color{red}{0}}^{\color{red}{1}}\frac{\log(1+t)}{1+t^2}\,dt$ we may simply invoke trigonometric identities. Such integral equals
$$ \int_{0}^{\pi/4}\log(1+\tan\theta)\,d\theta=\int_{0}^{\pi/4}\log(\sqrt{2}\sin\left(\theta+\tfrac{\pi}{4}\right))-\log(\sin\left(\tfrac{\pi}{2}-\theta\right))\,d\theta $$
which can be written as
$$ \int_{0}^{\pi/4}\frac{\log 2}{2}\,d\theta+\int_{\pi/4}^{\pi/2}\log\sin\theta\,d\theta-\int_{\pi/4}^{\pi/2}\log\sin\theta\,d\theta=\color{red}{\frac{\pi\log 2}{8}}.$$
By partial fraction decomposition, the primitive of $\frac{\log(1+t)}{1+t^2}$ depends on the dilogarithm/Spence's function, since $\int\frac{\log t}{a-t}\,dt=\log(t)\log\left(1-\frac{t}{a}\right)-\text{Li}_2\left(\frac{t}{a}\right)$:
$$ \int\frac{\log(1+t)}{(i+t)(-i+t)}\,dt = \frac{1}{2i}\int\log(1+t)\left(\frac{1}{-i-t}-\frac{1}{i-t}\right)\,dt. $$
Anyway, when dealing with integrals of involved functions, to remove the integration bounds is seldom a good idea, since it usually leads to more difficult problems by removing potentially useful symmetries.
A: HINT:
For solving $$\int \frac {ln(1+t)} {1+t^2} dt$$
You can substitute $t=tan\theta$, this will bring the integral down to
$$\int ln(1+tan\theta )d\theta$$
You can use by parts in this integration and this should be pretty much solvable.
