# How do you integrate int xsqrt(x+2)dx?

Nov 9, 2016

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

#### Explanation:

Let $= \int x \sqrt{x + 2} \mathrm{dx}$

We can integrate this easily by substitution:

Let $u = x + 2 \implies \frac{\mathrm{du}}{\mathrm{dx}} = 1$, or $\int \ldots \mathrm{du} = \int \ldots \mathrm{dx}$

Applying the substitution we have:

$I = \int \left(u - 2\right) \sqrt{u} \mathrm{du}$
$\therefore I = \int \left(u - 2\right) {u}^{\frac{1}{2}} \mathrm{du}$
$\therefore I = \int {u}^{\frac{3}{2}} - 2 {u}^{\frac{1}{2}} \mathrm{du}$
$\therefore I = {u}^{\frac{5}{2}} / \left(\frac{5}{2}\right) - 2 {u}^{\frac{3}{2}} / \left(\frac{3}{2}\right) + C$
$\therefore I = \frac{2}{5} {u}^{\frac{5}{2}} - \frac{4}{3} {u}^{\frac{3}{2}} + C$
$\therefore I = \frac{2}{5} {\left(x + 2\right)}^{\frac{5}{2}} - \frac{4}{3} {\left(x + 2\right)}^{\frac{3}{2}} + C$

We can put simplify by putting over a common denominator and factoring out ${\left(x + 2\right)}^{\frac{3}{2}}$

$\therefore I = \frac{6 {\left(x + 2\right)}^{\frac{5}{2}} - 20 {\left(x + 2\right)}^{\frac{3}{2}}}{15} + C$
$\therefore I = {\left(x + 2\right)}^{\frac{3}{2}} \frac{\left(6 \left(x + 2\right) - 20\right)}{15} + C$
$\therefore I = {\left(x + 2\right)}^{\frac{3}{2}} \frac{\left(6 x + 12 - 20\right)}{15} + C$
$\therefore I = {\left(x + 2\right)}^{\frac{3}{2}} \frac{\left(6 x - 8\right)}{15} + C$
$\therefore I = \frac{2 {\left(x + 2\right)}^{\frac{3}{2}} \left(3 x - 4\right)}{15} + C$