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

Apr 2, 2018

The integral is equal to $\frac{2}{3} \left({\left(x + 1\right)}^{\frac{3}{2}} + {x}^{\frac{3}{2}}\right) + C$.

#### Explanation:

First, rationalize the denominator:

$\textcolor{w h i t e}{=} \int \frac{1}{\sqrt{x + 1} - \sqrt{x}} \mathrm{dx}$

$= \int \frac{1}{\sqrt{x + 1} - \sqrt{x}} \textcolor{red}{\cdot \frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}}} \mathrm{dx}$

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

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

$= \int \frac{\sqrt{x + 1} + \sqrt{x}}{x + 1 - x} \mathrm{dx}$

$= \int \frac{\sqrt{x + 1} + \sqrt{x}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} + 1 \textcolor{red}{\cancel{\textcolor{b l a c k}{\textcolor{b l a c k}{-} x}}}} \mathrm{dx}$

$= \int \frac{\sqrt{x + 1} + \sqrt{x}}{1} \mathrm{dx}$

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

$= \int \sqrt{x + 1}$ $\mathrm{dx} + \int \sqrt{x}$ $\mathrm{dx}$

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

$= \frac{{\left(x + 1\right)}^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + \frac{{x}^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}$

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

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

You can factor out the $\frac{2}{3}$, and don't forget to add $C$:

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

That's the whole integral. Hope this helped!

Apr 2, 2018

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