I have the integral $$\int x\log(x)\ dx,$$ and I need to solve it by substitution. I don't know why, if you substitute $x$ by $g(y)$ you have $x= e^y$ and the integral of $e^y \cdot y \cdot e^y\ dy$ and the the integral of $e^{x^2}\cdot x\ dx$.
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$\begingroup$Try integration by parts, i.e. use:$$\int u\,dv=uv-\int v\,du$$ A prudent set of substitutions here would be: $$\begin{align} u&=\log(x)\\ dv&=x \, dx \end{align}$$ Hopefully you can finish this off
As noted by @SimonS below, you asked to do this by substitution alone. In which case try the substitution:$$u=2x^2\log(x)-x^2$$This leads to:$$du=4x\log(x)\,dx$$ Hopefully you can finish this off
$\endgroup$ 3 $\begingroup$There was an error in the arithmetic. Note that
$$e^y\,y\,e^y=ye^{2y}\ne ye^{y^2}$$
Therefore, we can write
$$\int x\log(x)\,dx=\int ye^{2y}\,dy$$
where $x=e^y$ or $y=\log(x)$.
$\endgroup$ 0 $\begingroup$Hint : $\ln x = z$ leads to $$\int x \ln(x) \, dx=\int z e^{2z} \, dz$$
The second integral can be easily calculated by integration by parts.
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