# What is the integral of int 1 / (sqrt(x+1) - sqrt(x)) ?

Feb 3, 2016

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

#### Explanation:

When I first looked at this one I thought it would be quite difficult, but in fact all we have to do is remember a basic technique from algebra: conjugate multiplication.

You may recall that the conjugate of a number is what you get when you switch the sign in the middle of a binomial. For example, the conjugate of $\sqrt{2} - 1$ is $\sqrt{2} + 1$ and the conjugate of $\sqrt{5} + \sqrt{3}$ is $\sqrt{5} - \sqrt{3}$. The interesting thing about conjugates is when you multiply them together, you get an interesting result. Look at the product of $\sqrt{2} - 1$ and $\sqrt{2} + 1$:
$\left(\sqrt{2} - 1\right) \left(\sqrt{2} + 1\right) = 2 + \sqrt{2} - \sqrt{2} - 1 = 2 - 1 = 1$.

We can see that even though our original expression contained square roots, our answer didn't - which, as we will see, is very useful.

In the denominator of our integral, we have $\sqrt{x + 1} - \sqrt{x}$. The conjugate of this is $\sqrt{x + 1} + \sqrt{x}$. Let's see what happens when we multiply our fraction by $\frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}}$ (note this is the same thing as multiplying by one):
$\int \frac{1}{\sqrt{x + 1} - \sqrt{x}} \mathrm{dx} = \int \frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}} \cdot \frac{1}{\sqrt{x + 1} - \sqrt{x}} \mathrm{dx}$

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

Doing some conjugate multiplication in the denominator,
$= \int \frac{\sqrt{x + 1} + \sqrt{x}}{\left(x + 1\right) - \left(\sqrt{x}\right) \left(\sqrt{x + 1}\right) + \left(\sqrt{x}\right) \left(\sqrt{x + 1}\right) - \left(x\right)} \mathrm{dx}$

The middle terms cancel out, leaving us with
$= \int \frac{\sqrt{x + 1} + \sqrt{x}}{\left(x + 1\right) - \left(x\right)} \mathrm{dx}$

And then the $x$s cancel out, so we have
$= \int \frac{\sqrt{x + 1} + \sqrt{x}}{1} \mathrm{dx}$

We see the beauty of conjugate multiplication now, as our seemingly complicated integral is now reduced to
$= \int \left(\sqrt{x + 1} + \sqrt{x}\right) \mathrm{dx}$

Using the properties of integrals,
$= \int \sqrt{x + 1} \mathrm{dx} + \int \sqrt{x} \mathrm{dx}$

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

Now all we have is a case of the reverse power rule in both integrals, making them simplify to
$\int \frac{1}{\sqrt{x + 1} - \sqrt{x}} \mathrm{dx} = \frac{2}{3} {\left(x + 1\right)}^{\frac{3}{2}} + \frac{2}{3} {\left(x\right)}^{\frac{3}{2}} + C$