# How do you integrate int x^5e^(-x^3)?

Nov 16, 2017

$\int {x}^{5} {e}^{- {x}^{3}} \mathrm{dx} = - \frac{{e}^{- {x}^{3}} \left({x}^{3} + 1\right)}{3} + C$

#### Explanation:

Substitute $t = {x}^{3}$ so that $\mathrm{dt} = 3 {x}^{2} \mathrm{dx}$ and note that:

${x}^{5} \mathrm{dx} = {x}^{3} / 3 \cdot 3 {x}^{2} \mathrm{dx} = \frac{t \mathrm{dt}}{3}$

We then have:

$\int {x}^{5} {e}^{- {x}^{3}} \mathrm{dx} = \frac{1}{3} \int t {e}^{- t} \mathrm{dt}$

Now integrate by parts using $t$ as integer factor:

$\int t {e}^{- t} \mathrm{dt} = - \int t d \left({e}^{- t}\right) = - t {e}^{- t} + \int {e}^{- t} \mathrm{dt} = - {e}^{- t} \left(t + 1\right) + C$

and undoing the substitution:

$\int {x}^{5} {e}^{- {x}^{3}} \mathrm{dx} = - \frac{{e}^{- {x}^{3}} \left({x}^{3} + 1\right)}{3} + C$