# How do you integrate by substitution int x^2(x^3+5)^4 dx?

Oct 31, 2016

$\int {x}^{2} {\left({x}^{3} + 5\right)}^{4} \mathrm{dx} = {\left({x}^{3} + 5\right)}^{5} / 15 + C$

#### Explanation:

We want to find $I = \int {x}^{2} {\left({x}^{3} + 5\right)}^{4} \mathrm{dx}$

Let $u = {x}^{3} + 5 \implies \frac{\mathrm{du}}{\mathrm{dx}} = 3 {x}^{2}$, or $3 {x}^{2} \frac{\mathrm{dx}}{\mathrm{du}} = 1$

We can then rewrite $I$ as follows:

$I = \int {x}^{2} {u}^{4} \mathrm{dx}$
$\therefore I = \frac{1}{3} \int {u}^{4} \left(3 {x}^{2}\right) \mathrm{dx}$
$\therefore I = \frac{1}{3} \int {u}^{4} \left(3 {x}^{2}\right) \frac{\mathrm{dx}}{\mathrm{du}} \mathrm{du}$ (by the chain rule)

And using the above result we can now substitute to get:
$I = \frac{1}{3} \int {u}^{4} \left(1\right) \mathrm{du}$
$\therefore I = \frac{1}{3} \int {u}^{4} \mathrm{du}$
$\therefore I = \frac{1}{3} {u}^{5} / 5 + C$
$\therefore I = {\left({x}^{3} + 5\right)}^{5} / 15 + C$