# wanna confirm ans (1 Viewer)

#### vishnay

##### God
i got $\bg_white \frac{x^2e^{x^2}}{2}-\frac{e^{x^2}}{2}+C$

#### 5uckerberg

##### Well-Known Member
Is this integration by parts.
What is the differentiation of $\bg_white e^{x^{2}}$? Knowing that we can now have
$\bg_white \int{x^{3}e^{x^{2}}}dx=\frac{1}{2}\int{2x^{3}e^{x^{2}}}dx=\frac{1}{2}\int{x^{2}\cdot{2xe^{x^{2}}}}dx$.
There, we will have $\bg_white \frac{1}{2}\int{x^{2}\cdot{2xe^{x^{2}}}}dx=\frac{1}{2}x^{2}e^{x^{2}}-\int{xe^{x^{2}}}dx$
Note that $\bg_white \int{xe^{x^{2}}}dx=\frac{1}{2}\int{2xe^{x^{2}}}dx$. So our final answer should be
$\bg_white \frac{e^{x^{2}}}{2}\left(x^{2}-1\right)$