$\int x^{3} \cdot e^{x^{2}} d x=$

Question

Aicharya · Accepted Answer

Let $I=\int x^{3} e^{x^{2}} d x$ Put $x^{2}=t \Rightarrow 2 x d x=d t \Rightarrow x d x=\frac{1}{2} d t$ $	herefore \mathrm{I}=\frac{1}{2} \int \mathrm{te}^{\mathrm{t}} \mathrm{dt}$ $=\frac{1}{2}\left[t e^{t}-e^{t}ight]+c=\frac{1}{2} e^{t}(t-1)+c=\frac{1}{2} e^{x^{2}}\left(x^{2}-1ight)+c$

$\int x^{3} \cdot e^{x^{2}} d x=$

Solution