# How do you integrate e^7x^3 x^2 dx?

Aug 15, 2016

$= \frac{{e}^{7}}{6} \left({x}^{6}\right) + C$

#### Explanation:

$\int {e}^{7} {x}^{3} {x}^{2} \setminus \mathrm{dx}$

I can see two approaches 1) combining the $x$ terms or 2) doing u substitution. first lets express this differently by moving out the constant

${e}^{7} \int {x}^{3} {x}^{2} \setminus \mathrm{dx}$
1)

${e}^{7} \int {x}^{5} \setminus \setminus \mathrm{dx}$
$\frac{{e}^{7}}{6} \left({x}^{6}\right) + C$

2)
now i will use u substitution and let $u = {x}^{3}$ then
$\frac{\mathrm{du}}{\mathrm{dx}} = 3 {x}^{2}$
$\frac{1}{3} \mathrm{du} = {x}^{2} \mathrm{dx}$

now substituting this back in
${e}^{7} \int u \frac{1}{3} \setminus \setminus \mathrm{du}$
$\frac{{e}^{7}}{3} \int u \setminus \setminus \mathrm{du}$

$= \frac{{e}^{7}}{3} \frac{{u}^{2}}{2} + C$
now we place replace u
$= \frac{{e}^{7}}{6} {\left({x}^{3}\right)}^{2} + C$
$= \frac{{e}^{7}}{6} \left({x}^{6}\right) + C$