# What is the antiderivative of x sqrtx?

Jan 17, 2017

$\frac{2}{5} {x}^{2} \sqrt{x} + C$

#### Explanation:

Note that $x \sqrt{x} = {x}^{1} {x}^{\frac{1}{2}} = {x}^{1 + \frac{1}{2}} = {x}^{\frac{3}{2}}$. Then:

$\int x \sqrt{x} \textcolor{w h i t e}{.} \mathrm{dx} = \int {x}^{\frac{3}{2}} \mathrm{dx}$

Now use the rule $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C$ we see that:

$= {x}^{\frac{3}{2} + 1} / \left(\frac{3}{2} + 1\right) + C = {x}^{\frac{5}{2}} / \left(\frac{5}{2}\right) + C = \frac{2}{5} {x}^{\frac{5}{2}} + C$

If you so desire, you can write that

${x}^{\frac{5}{2}} = {x}^{\frac{4}{2}} {x}^{\frac{1}{2}} = {x}^{2} \sqrt{x}$

So:

$\int x \sqrt{x} \textcolor{w h i t e}{.} \mathrm{dx} = \frac{2}{5} {x}^{2} \sqrt{x} + C$