# How do you integrate int xsqrt(x-1) by parts?

Jan 30, 2017

$x = {\cos}^{2} y$
$\sqrt{x - 1} = \sin y$
$I = \int {\cos}^{2} y \sin y 2 \cos y \sin y \left(- 1\right)$
$I = \int \left(- 2\right) {\cos}^{3} y \sin y$
$I = \frac{1}{2} {\cos}^{4} y$

Jan 30, 2017

#### Explanation:

$\int x \sqrt{x - 1} \mathrm{dx}$

Let $u = x$ and $\mathrm{dv} = \sqrt{x - 1} \mathrm{dx}$

so that $\mathrm{du} = 1 \mathrm{dx}$ and $v = \frac{2}{3} {\left(x - 1\right)}^{\frac{3}{2}}$

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

$= \frac{2}{3} x {\left(x - 1\right)}^{\frac{3}{2}} - \frac{2}{3} \left[\frac{2}{5} {\left(x - 1\right)}^{\frac{5}{2}}\right] + C$

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

Rewrite algebraically to taste. I like the answer above, but others might prefer

$= \frac{2}{15} \left[5 x {\left(x - 1\right)}^{\frac{3}{2}} - 2 {\left(x - 1\right)}^{\frac{5}{2}}\right] + C$

Or

$= \frac{2}{15} {\left(x - 1\right)}^{\frac{3}{2}} \left(3 x + 2\right) + C$

Or some equivalent expression.

Jan 30, 2017

Without integration by parts one can see that
$x = x - 1 + 1$
so
$x \sqrt{x - 1} = \left(x - 1\right) \sqrt{x - 1} + \sqrt{x - 1}$
$\int x \sqrt{x - 1} \mathrm{dx} = \int {\left(x - 1\right)}^{\frac{3}{2}} d \left(x - 1\right) + \int {\left(x - 1\right)}^{\frac{1}{2}} d \left(x - 1\right)$
$\int x \sqrt{x - 1} \mathrm{dx} = {\left(x - 1\right)}^{\frac{5}{2}} \frac{2}{5} + {\left(x - 1\right)}^{\frac{3}{2}} \frac{2}{3}$