evaluate $\int \frac{x^2\cdot\log(x)}{x+1} dx$ I am trying to evaluate integral:      
$$\int \frac{x^2\log(x)}{x+1}dx$$
But I have some problems with it. If I use  Wolfram Alpha like this
I get a result, but I need evaluate it by hand. Which method should I use?
If we represent it as:
$$\int(x^2\log(x)\cdot d(\ln(x))dx$$
then it is unclear what to dot here, even we can represent as power, take $x^2$ in power of $\log(x)$, not hint yet. Please help me to evaluate it.

We have 
\begin{align*}
\int\frac{x^2\log x}{x+1}dx&=\int x\log x\frac{x+1-1}{x+1}dx\\
&=\int x\log x-\int \frac{x+1-1}{x+1}\log xdx\\
&=\frac{x^2}2\log x-\int\frac{x^2}2\frac 1xdx-\int \log x+\int\frac{\log x}{x+1}dx\\
&=\frac{x^2}2\log x-\frac{x^2}4-x\log x+x+\log x\log(x+1)-\int\frac{\log(x+1)}xdx\\
&=\frac x4(4-x)+\log x\left(\frac{x^2}2-x+\log(x+1)\right)-\int\frac{\log(x+1)}xdx\\
&=\frac x4(4-x)+\log x\left(\frac{x^2}2-x+\log(x+1)\right)-\sum_{k=1}^{+\infty}\int \frac{(-1)^{k+1}x^k}{kx}dx\\
&=\frac x4(4-x)+\log x\left(\frac{x^2}2-x+\log(x+1)\right)+\sum_{k=1}^{+\infty}(-1)^k \frac{x^k}{k^2}\\
&=\frac x4(4-x)+\log x\left(\frac{x^2}2-x+\log(x+1)\right)+\operatorname{Li}_2(-x),
\end{align*}
which is the result given by Wolfram Alpha.
