# Question #a0e4b

May 19, 2017

$\left(\frac{1}{4}\right) \ln \left(\left\mid x - 2 \right\mid\right) - \left(\frac{1}{4}\right) \left(\ln \left(\left\mid x + 2 \right\mid\right)\right) + C$

#### Explanation:

I will assume you meant $\int \left(\frac{\mathrm{dx}}{{x}^{2} - 4}\right)$:

Rewrite ${x}^{2} - 4$ as $\left(x - 2\right) \left(x + 2\right)$:

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

Rewrite the fraction as $\frac{A}{x - 2} + \frac{B}{x + 2}$. Solve for $A$ and $B$ by multiplying the three fractions by $\left(x - 2\right) \left(x + 2\right)$:

$1 = A \left(x + 2\right) + B \left(x - 2\right)$

Plug in $2$ for $x$ to solve for $A$:

$1 = A \left(4\right)$

$A = \frac{1}{4}$

Plug in $- 2$ for $x$ to solve for $B$:

$1 = - 4 B$

$B = - \frac{1}{4}$

Now we have:

$\int \left(\frac{1}{4}\right) \left(\frac{1}{x - 2}\right) + \left(- \frac{1}{4}\right) \left(\left(\frac{1}{x + 2}\right)\right)$

This, we can solve by integrating:

$\left(\frac{1}{4}\right) \ln \left(\left\mid x - 2 \right\mid\right) - \left(\frac{1}{4}\right) \left(\ln \left(\left\mid x + 2 \right\mid\right)\right) + C$

If you want, you can write this as one log with log rules:

$\left(\frac{1}{4}\right) \ln \left(\left\mid \frac{x - 2}{x + 2} \right\mid\right) + C$