Find $\int x\ln(x^2e^{x^2})\,dx.$ How can I find the following integral $$\int x\ln(x^2e^{x^2})\,dx.$$ Which substitution may I use to solve the integral?
 A: Notice $$\ln(x^2 e^{x^2}) = \ln(x^2)+\ln(e^{x^2}) = \ln(x^2)+x^2$$
so
$$\int x \ln(x^2 e^{x^2})\text{ d}x = \int x[\ln(x^2)+x^2]\text{ d}x = \int x\ln(x^2)\text{ d}x + \int x^3\text{ d}x$$
The second integral is easy; for the first integral, let $u = x^2$. Then use the fact that $$\int\ln(x)\text{ d}x = x\ln(x)-x+C\text{.}\tag{1}$$
If you haven't proven $(1)$, using integration by parts with $u = \ln(x)$ and $\text{d}v = \text{d}x$ gives $\text{d}u = \dfrac{1}{x}\text{ d}x$ and $v = x$, so
$$\int\ln(x)\text{ d}x = uv-\int v\text{ d}u =  x\ln(x) - \int\dfrac{x}{x}\text{ d}x = x\ln(x)-x+C\text{.}$$
A: This looks much simpler:
 $$\int x\ln(x^2e^{x^2})dx =\int x\left(\ln(x^2) + \ln(e^{x^2})\right)dx =\int x\left(2\ln(x) + {x^2}\right)dx =\int 2x\ln(x) + {x^3}dx $$ $$ = 2\int x\ln(x)dx + \int {x^3}dx  $$
A: Note that by applying logarithms addition rule you get
$$\begin{align}
\int x\ln(x^2e^{x^2})dx &= \int x\left(\ln(x^2) + \ln(e^{x^2})\right)dx\\
&=\int x\left(2\ln x  + x^2\right)dx\\
&=\int 2x\ln x + x^3dx\\
&= 2\int x\ln x + \int x^3dx\\
&= 2\left( \frac{1}{2} x^2 \log (x)-\frac{x^2}{4} \right) + \frac{x^4}{4} + C\\
&= x^2 \log x - \frac{x^2}{2} + \frac{x^4}{4}
\end{align}$$
