Q:

How do I find the closed form of this integral $\int_0^2\frac{\ln x}{x^3-2x+4}dx$?

How do I find the closed form of this integral:
$$I=\int_0^2\frac{\ln x}{x^3-2x+4}dx$$
First, I have a partial fraction of it:
$$\frac{1}{x^3-2x+4}=\frac{1}{(x+2)(x^2-2x+2)}=\frac{A}{x+2}+\frac{Bx+C}{x^2-2x+2}$$
$$A=\frac{1}{(x^3-2x+4)'}|_{x=-2}=\frac{1}{(3x^2-2)}|_{x=-2}=\frac{1}{10}$$
$$Bx+C=\frac{1}{x+2}|_{x^2-2x=-2}=\frac{1}{(x+2)}\frac{(x-4)}{(x-4)}|_{x^2-2x=-2}=$$
$$=\frac{(x-4)}{(x^2-2x-8)}|_{x^2-2x=-2}=\frac{(x-4)}{(-2-8)}|_{x^2-2x=-2}=-\frac{1}{10}(x-4)$$
Thus:
$$\frac{1}{x^3-2x+4}=\frac{1}{10}\left(\frac{1}{x+2}-\frac{x-4}{x^2-2x+2}\right)$$
$$I=\frac{1}{10}\left(\int_0^2\frac{\ln x}{x+2}dx-\int_0^2\frac{(x-4)\ln x}{x^2-2x+2}dx\right)$$
What should I do next?

A:

$\displaystyle I=\int_0^2\frac{\ln x}{x^3-2x+4}\,dx$
$\begin{align}I=\frac{1}{10}\int_0^2 \frac{\ln x}{2+x}\,dx+\frac{4}{10}\int_0^2 \frac{\ln x}{x^2-2x+2}\,dx-\frac{1}{10}\int_0^2 \frac{x\ln x}{x^2-2x+2}\,dx\end{align}$
$\begin{align}
A&:=\int_0^2 \frac{\ln x}{2+x}\,dx\\
&=\frac{1}{2}\int_0^2 \frac{\ln x}{1+\frac{x}{2}}\,dx\\
\end{align}$
Perform the change of variable $\displaystyle y=\frac{x}{2}$,
$\begin{align}
A&=\int_0^1 \frac{\ln(2y)}{1+y}\,dy\\
&=\int_0^1 \frac{\ln 2}{1+y}\,dy+\int_0^1 \frac{\ln y}{1+y}\,dy\\
&=\ln 2\Big[\ln(1+y)\Big]_0^1-\frac{\pi^2}{12}\\
&=\boxed{\ln^2 2-\frac{\pi^2}{12}}
\end{align}$
It is well-known that,
$\begin{align}\int_0^1 \frac{\ln y}{1+y}\,dy=-\frac{1}{2}\zeta(2)\end{align}$
and,
$\displaystyle \zeta(2)=\frac{\pi^2}{6}$
$\begin{align}B&:=\int_0^2 \frac{\ln x}{x^2-2x+2}\,dx\\
&=\int_0^2\frac{\ln x}{(x-1)^2+1}\,dx
\end{align}$
Perform the change of variable $y=x-1$,
$\begin{align}B&=\int_{-1}^1\frac{\ln(1+y)}{y^2+1}\,dy\\
&=\int_{-1}^0\frac{\ln(1+y)}{y^2+1}\,dy+\int_{0}^1\frac{\ln(1+y)}{y^2+1}\,dy
\end{align}$
Perform the change of variable $x=-y$ in the first integral,
$\begin{align}B&=\int_{0}^1\frac{\ln(1-y)}{y^2+1}\,dy+\int_{0}^1\frac{\ln(1+y)}{y^2+1}\,dy\\
\end{align}$
In the first integral perform the change of variable $\displaystyle x=\frac{1-y}{1+y}$
$\begin{align}B&=\left(\int_0^1\frac{\ln 2}{1+x^2}\,dx+\int_0^1\frac{\ln x}{1+x^2}\,dx-\int_0^1\frac{\ln(1+x)}{1+x^2}\,dx\right)+\int_{0}^1\frac{\ln(1+y)}{1+y^2}\,dy\\
&=\boxed{\frac{1}{4}\pi\ln 2-\text{G}}
\end{align}$
It is well-known that,
$\displaystyle \int_0^1 \frac{\ln x}{1+x^2}\,dx=-\text{G}$
$\begin{align}C&:=\int_0^2 \frac{x\ln x}{x^2-2x+2}\,dx\\
&=\int_0^2\frac{x\ln x}{(x-1)^2+1}\,dx
\end{align}$
Perform the change of variable $y=x-1$,
$\begin{align}C&=\int_{-1}^1\frac{(1+y)\ln(1+y)}{y^2+1}\,dy\\
&=B+\int_{-1}^1\frac{y\ln(1+y)}{y^2+1}\,dy\\
&=\frac{1}{4}\pi\ln 2-\text{G}+\int_{-1}^0\frac{y\ln(1+y)}{y^2+1}\,dy+\int_{0}^1\frac{y\ln(1+y)}{y^2+1}\,dy
\end{align}$
In the first integral perform the change of variable $x=-y$,
$\begin{align}C&=\frac{1}{4}\pi\ln 2-\text{G}-\int_{0}^1\frac{x\ln(1-x)}{x^2+1}\,dx+\int_{0}^1\frac{x\ln(1+x)}{x^2+1}\,dx\\
&=\frac{1}{4}\pi\ln 2-\text{G}+\int_0^1 \frac{x\ln\left(\frac{1+x}{1-x}\right)}{1+x^2}\,dx
\end{align}$
Perform the change of variable $\displaystyle y=\frac{1-x}{1+x}$,
$\begin{align}C&=\frac{1}{4}\pi\ln 2-\text{G}+\int_0^1 \frac{(x-1)\ln x}{x^3+x^2+x+1}\,dx\\
&=\frac{1}{4}\pi\ln 2-\text{G}+\int_0^1\frac{x\ln x}{1+x^2}\,dx-\int_0^1\frac{\ln x}{1+x}\,dx\\
&=\frac{1}{4}\pi\ln 2-\text{G}+\frac{1}{4}\int_0^1\frac{2x\ln(x^2)}{1+x^2}\,dx-\int_0^1\frac{\ln x}{1+x}\,dx\\
\end{align}$
In the first integral perform the change of variable $y=x^2$,
$\begin{align}C&=\frac{1}{4}\pi\ln 2-\text{G}+\frac{1}{4}\int_0^1\frac{\ln x}{1+x}\,dx-\int_0^1\frac{\ln x}{1+x}\,dx\\
&=\frac{1}{4}\pi\ln 2-\text{G}-\frac{3}{4}\int_0^1\frac{\ln x}{1+x}\,dx\\
&=\boxed{\frac{1}{4}\pi\ln 2-\text{G}+\frac{1}{16}\pi^2}
\end{align}$
Therefore,
$\boxed{\displaystyle I=\frac{1}{10}\ln^2 2+\frac{3}{40}\pi\ln 2-\frac{3}{10}\text{G}-\frac{7}{480}\pi^2}$

