# How do you integrate int (x+1)sqrt(2-x)dx?

Dec 10, 2016

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

#### Explanation:

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

Let $u = 2 - x$. This implies that $\mathrm{du} = - \mathrm{dx}$. Also note that $- u = x - 2$, so $- u + 3 = x + 1$. Then:

$I = \int \left(- u + 3\right) \sqrt{u} \left(- \mathrm{du}\right) = \int \left(u - 3\right) \sqrt{u} \mathrm{du}$

Expanding the square root as ${u}^{\frac{1}{2}}$:

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

Now using $\int {u}^{n} \mathrm{du} = {u}^{n + 1} / \left(n + 1\right) + C$:

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

Factoring and making it look nice:

$I = {u}^{\frac{3}{2}} \left(\frac{2}{5} u - 2\right) = \frac{{u}^{\frac{3}{2}} \left(2 u - 10\right)}{5} = \frac{2 {u}^{\frac{3}{2}} \left(u - 5\right)}{5}$

From $u = 2 - x$:

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