# How do you find the integral of f(x)=xsqrt(x-5) using integration by parts?

Nov 16, 2015

$\frac{2}{3} x {\left(x - 5\right)}^{\frac{3}{2}} - \frac{4}{15} {\left(x - 5\right)}^{\frac{5}{2}}$

#### Explanation:

First, here's the formula for "integration by parts":

$\int f ' \left(x\right) \cdot g \left(x\right) \text{d"x = f(x) g(x) - int f(x) g'(x) "d} x$

Now, in your product $x \cdot \sqrt{x - 5}$ you need to determine which factor is $f ' \left(x\right)$ and which one is $g \left(x\right)$ since you will need to integrate one of them and differentiate the other one.

As it doesn't make a big difference with the radical expression but would be a good thing to differentiate $x$, let's assume that $f \left(x\right) = x$ and $g ' \left(x\right) = \sqrt{x - 5}$.

As next, you need to differentiate $f \left(x\right)$ and integrate $g ' \left(x\right)$:
$f \left(x\right) = x \implies f ' \left(x\right) = 1$
$g ' \left(x\right) = \sqrt{x - 5} = {\left(x - 5\right)}^{\frac{1}{2}} \implies g \left(x\right) = \frac{2}{3} \cdot {\left(x - 5\right)}^{\frac{3}{2}}$

Now we can use the formula:

$\int x \cdot \sqrt{x - 5} \text{d} x$
$= x \cdot \frac{2}{3} \cdot {\left(x - 5\right)}^{\frac{3}{2}} - \int 1 \cdot \frac{2}{3} \cdot {\left(x - 5\right)}^{\frac{3}{2}} \text{d} x$
$= \frac{2}{3} x {\left(x - 5\right)}^{\frac{3}{2}} - \frac{2}{3} \int {\left(x - 5\right)}^{\frac{3}{2}} \text{d} x$
$= \frac{2}{3} x {\left(x - 5\right)}^{\frac{3}{2}} - \frac{2}{3} \cdot \frac{2}{5} {\left(x - 5\right)}^{\frac{5}{2}}$
$= \frac{2}{3} x {\left(x - 5\right)}^{\frac{3}{2}} - \frac{4}{15} {\left(x - 5\right)}^{\frac{5}{2}}$

Hope that this helped!