# How do you evaluate the integral int x/(sqrt(x-1)-sqrtx)?

Jan 29, 2017

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

#### Explanation:

Multiply first by the conjugate of the denominator to simplify the integrand.

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

The denominator is now in the form $\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$, where $a = \sqrt{x - 1}$ and $b = x$.

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

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

Distributing, splitting up the integral and rewriting:

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

The second integral can be directly integrated using $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C$, where $n \ne - 1$.

$I = - \int x {\left(x - 1\right)}^{\frac{1}{2}} \mathrm{dx} - {x}^{\frac{5}{2}} / \left(\frac{5}{2}\right)$

$I = - \int x {\left(x - 1\right)}^{\frac{1}{2}} \mathrm{dx} - \frac{2}{5} {x}^{\frac{5}{2}}$

For the remaining integral, let $u = x - 1$. This implies that $x = u + 1$ and $\mathrm{du} = \mathrm{dx}$.

$I = - \int \left(u + 1\right) {u}^{\frac{1}{2}} \mathrm{du} - \frac{2}{5} {x}^{\frac{5}{2}}$

Distributing $\left(u + 1\right) {u}^{\frac{1}{2}}$ into separate integrals:

$I = - \int {u}^{\frac{3}{2}} \mathrm{du} - \int {u}^{\frac{1}{2}} \mathrm{du} - \frac{2}{5} {x}^{\frac{5}{2}}$

Using the rule from earlier:

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

From $u = x - 1$:

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