# How do you integrate int 1/(x^2-x-20) dx using partial fractions?

May 7, 2018

$\frac{1}{9} \ln | x - 5 | - \frac{1}{9} \ln | x + 4 | + C$

#### Explanation:

Factor the denominator of the integrand:

${x}^{2} - x - 20 = \left(x - 5\right) \left(x + 4\right)$

Use partial fraction decomposition on the simplified integrand:

$\frac{1}{\left(x - 5\right) \left(x + 4\right)} = \frac{A}{x - 5} + \frac{B}{x + 4}$

$\frac{1}{\left(x - 5\right) \left(x + 4\right)} = \frac{A \left(x + 4\right) + B \left(x - 5\right)}{\left(x - 5\right) \left(x + 4\right)}$

Equate numerators:

$1 = A \left(x + 4\right) + B \left(x - 5\right)$

We need to find $A , B .$ We can do this by plugging in values of $x$ that send one term to $0$ and keep the other:

$x = - 4 :$

$1 = - 9 B , B = - \frac{1}{9}$

$x = 5 :$

$1 = 9 A , A = \frac{1}{9}$

Thus, our integral becomes

$\int \frac{\frac{1}{9}}{x - 5} - \frac{\frac{1}{9}}{x + 4} \mathrm{dx} = \frac{1}{9} \ln | x - 5 | - \frac{1}{9} \ln | x + 4 | + C$

$\frac{1}{9} \cdot \ln \left\mid x - 5 \right\mid - \frac{1}{9} \cdot \ln \left\mid x + 4 \right\mid$

#### Explanation:

$\frac{1}{\left(x - 5\right) \left(x + 4\right)} = \frac{1}{9 \cdot \left(x - 5\right)} - \frac{1}{9 \cdot \left(x + 4\right)}$

$\int \frac{1}{\left(x - 5\right) \left(x + 4\right)} = \int \frac{1}{9 \cdot \left(x - 5\right)} - \frac{1}{9 \cdot \left(x + 4\right)} \mathrm{dx}$

$\int \frac{1}{\left(x - 5\right) \left(x + 4\right)} = \int \frac{1}{9 \cdot \left(x - 5\right)} - \frac{1}{9 \cdot \left(x + 4\right)} \mathrm{dx}$

$\int \frac{1}{\left(x - 5\right) \left(x + 4\right)} = \frac{1}{9} \cdot \ln \left\mid x - 5 \right\mid - \frac{1}{9} \cdot \ln \left\mid x + 4 \right\mid$