# Question #56aaf

##### 1 Answer
Feb 14, 2017

$\int \left(x - 1\right) \sqrt{2 + x} \mathrm{dx} = \left(x - 1\right) \frac{2}{3} {\left(2 + x\right)}^{\frac{3}{2}} + \frac{4}{15} {\left(2 + x\right)}^{\frac{5}{2}} + C$

#### Explanation:

Let $u = x - 1$ and $\mathrm{dv} = \sqrt{2 + x} \mathrm{dx}$. By the product rule, $\mathrm{du} = \mathrm{dx}$. To integrate $\mathrm{dv}$ however, we will need to make a substitution.

Let $n = 2 + x$. Then $\mathrm{dn} = \mathrm{dx}$.

$\mathrm{dv} = \sqrt{n} \mathrm{dn}$

$\int \mathrm{dv} = \int \sqrt{n} \mathrm{dn}$

$v = \frac{2}{3} {n}^{\frac{3}{2}} \to$ We can add $C$ later on

$v = \frac{2}{3} {\left(2 + x\right)}^{\frac{3}{2}}$

Now use the integration by parts formula.

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

$\int \left(x - 1\right) \sqrt{2 + x} \mathrm{dx} = \left(x - 1\right) \frac{2}{3} {\left(2 + x\right)}^{\frac{3}{2}} - \int \left(\frac{2}{3} {\left(2 + x\right)}^{\frac{3}{2}}\right) \mathrm{dx}$

You're going to need substitution to evaluate this integral. Let $m = 2 + x$. Then $\mathrm{dm} = \mathrm{dx}$.

$\int \frac{2}{3} {\left(2 + x\right)}^{\frac{3}{2}} \mathrm{dx} = \int \frac{2}{3} {m}^{\frac{3}{2}} \mathrm{dm}$

Now use $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C$, as we did above.

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

Reverse the substitution:

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

We can now put everything together.

$\int \left(x - 1\right) \sqrt{2 + x} \mathrm{dx} = \left(x - 1\right) \frac{2}{3} {\left(2 + x\right)}^{\frac{3}{2}} + \frac{4}{15} {\left(2 + x\right)}^{\frac{5}{2}} + C$

Hopefully this helps!