# How do you integrate (x^2-1)/(x^2-16) using partial fractions?

Dec 2, 2016

The answer is =x-15/8ln(∣x+4∣)+15/8ln(∣x-4∣)+C

#### Explanation:

Firstly

$\frac{{x}^{2} - 1}{{x}^{2} - 16} = 1 + \frac{15}{{x}^{2} - 16}$

${x}^{2} - 16 = \left(x + 4\right) \left(x - 1\right)$

Now we can do the decomposition in partial fractions

$\frac{15}{{x}^{2} - 16} = \frac{15}{\left(x + 4\right) \left(x - 1\right)} = \frac{A}{x + 4} + \frac{B}{x - 4}$

$= \frac{A \left(x - 4\right) + B \left(x + 4\right)}{\left(x + 4\right) \left(x - 1\right)}$

Therefore,

$15 = A \left(x - 4\right) + B \left(x + 4\right)$

Let $x = 4$, $\implies$, $15 = 8 B$

$B = \frac{15}{8}$

Let $x = - 4$, $\implies$, $15 = - 8 A$

$A = - \frac{15}{8}$

(x^2-1)/(x^2-16)=1-15/(8(x+4))+15/(8(x-4)

int((x^2-1)dx)/(x^2-16)=int1dx-int(15dx)/(8(x+4))+int(15dx)/(8(x-4)

=x-15/8ln(∣x+4∣)+15/8ln(∣x-4∣)+C