# How do you evaluate the integral int xsqrt(x-2)?

Jan 29, 2017

The answer is $= \frac{2}{15} {\left(x - 2\right)}^{\frac{3}{2}} \left(3 x + 4\right) + C$

#### Explanation:

We need

$\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C \left(x \ne - 1\right)$

We solve this integral by substitution

Let $u = x - 2$

$\mathrm{du} = \mathrm{dx}$

and $x = u + 2$

Therefore,

$\int x \sqrt{x - 2} \mathrm{dx} = \int \left(u + 2\right) \sqrt{u} \mathrm{du}$

$= \int \left({u}^{\frac{3}{2}} + 2 {u}^{\frac{1}{2}}\right) \mathrm{du}$

$= {u}^{\frac{5}{2}} / \left(\frac{5}{2}\right) + 2 \cdot {u}^{\frac{3}{2}} / \left(\frac{3}{2}\right)$

$= \frac{2}{5} {u}^{\frac{5}{2}} + \frac{4}{3} {u}^{\frac{3}{2}}$

$= \frac{2}{15} {u}^{\frac{3}{2}} \left(3 u + 10\right)$

$= \frac{2}{15} {\left(x - 2\right)}^{\frac{3}{2}} \left(3 x - 6 + 10\right) + C$

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