# What is int1/(x^2-5)?

## When I use an online integral calculator, it's suggesting I use substitution and let $u = \frac{x}{\sqrt{5}}$ I don't understand how to get the $\sqrt{5}$. Is there an alternative way of solving this? If not - what does the $\sqrt{5}$ mean and how do I work this out for similar questions?

May 16, 2017

$\ln \frac{\left\mid \frac{x - \sqrt{5}}{x + \sqrt{5}} \right\mid}{2 \sqrt{5}} + C$

#### Explanation:

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

Now if we set $u = \frac{x}{\sqrt{5}}$ then the numerator must be $\frac{1}{\sqrt{5}}$ since $\frac{\mathrm{du}}{\mathrm{dx}} \left[u\right] = \frac{1}{\sqrt{5}}$ to apply the substitition rule.

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

It is now all set for substitution:
$\frac{\sqrt{5}}{5} \int \frac{1}{{u}^{2} - 1} \mathrm{du} = \left(\frac{\sqrt{5}}{5}\right) \left[\frac{1}{2} \left(\int \frac{1}{u - 1} \mathrm{du} - \int \frac{1}{u + 1} \mathrm{du}\right)\right] = \left(\frac{1}{2 \sqrt{5}} \left[\ln \left(u - 1\right) - \ln \left(u + 1\right)\right]\right) = \frac{\ln \left[\frac{u - 1}{u + 1}\right]}{2 \sqrt{5}}$

Undo substitution:
$\frac{\ln \left[\frac{\frac{x}{\sqrt{5}} - 1}{\frac{x}{\sqrt{5}} + 1}\right]}{2 \sqrt{5}} = \ln \frac{\left\mid \frac{x - \sqrt{5}}{x + \sqrt{5}} \right\mid}{2 \sqrt{5}} + C$

Absolute value for the domain.

I use to solve this way: