How to integrate $\int{\frac{\ln(1+t)}{t^2+1}}dt$

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}