# How do you use substitution to integrate (r^2) √(r^(3) +3)dr?

##### 1 Answer
Mar 25, 2018

$\int {r}^{2} / \sqrt{{r}^{3} + 3} \mathrm{dr} = \frac{2}{3} \sqrt{{r}^{3} + 3} + C$

#### Explanation:

So, we want to determine

$\int {r}^{2} / \sqrt{{r}^{3} + 3} \mathrm{dr}$

The substitution should be picked in such a way that the differential of what we've picked will show up in the integral.

The best choice is $u = {r}^{3} + 3$, as

$\mathrm{du} = 3 {r}^{2} \mathrm{dr}$

$3 {r}^{2} \mathrm{dr}$ does not show up in this exact form in the integral; however, $\frac{1}{3} \mathrm{du} = {r}^{2} \mathrm{dr}$ shows up in the numerator of the integral. Thus, we can substitute and now have

$\frac{1}{3} \int \frac{\mathrm{du}}{\sqrt{u}} = \frac{1}{3} \int {u}^{- \frac{1}{2}} \mathrm{du}$ (We've factored $\frac{1}{3}$ outside of the integral)

Integrate:

$\left(\frac{1}{3}\right) \left(2\right) \sqrt{u} + C$

Rewriting in terms of $x$ yields

$\int {r}^{2} / \sqrt{{r}^{3} + 3} \mathrm{dr} = \frac{2}{3} \sqrt{{r}^{3} + 3} + C$