# How do you integrate int(x+1)/((x+5)(x-1)(x-4)) using partial fractions?

Sep 1, 2017

The integral is $= - \frac{2}{27} \ln \left(| x + 5 |\right) - \frac{1}{9} \ln \left(| x - 1 |\right) + \frac{5}{27} \ln \left(| x - 4 |\right) + C$

#### Explanation:

Let's perform the decomposition into partial fractions

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

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

The denominators are the same, we compare the numerators

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

Let $x = - 5$, $\implies$, $- 4 = A \cdot \left(- 6\right) \left(- 9\right)$, $\implies$, $A = - \frac{2}{27}$

Let $x = 1$, $\implies$, $2 = B \cdot \left(6\right) \left(- 3\right)$, $\implies$, $B = - \frac{1}{9}$

Let $x = 4$, $\implies$, $5 = C \cdot \left(9\right) \left(3\right)$, $\implies$, $C = \frac{5}{27}$

Therefore,

$\frac{x + 1}{\left(x + 5\right) \left(x - 1\right) \left(x - 4\right)} = \frac{- \frac{2}{27}}{x + 5} + \frac{- \frac{1}{9}}{x - 1} + \frac{\frac{5}{27}}{x - 4}$

$\int \frac{\left(x + 1\right) \mathrm{dx}}{\left(x + 5\right) \left(x - 1\right) \left(x - 4\right)} = \int \frac{- \frac{2}{27} \mathrm{dx}}{x + 5} + \int \frac{- \frac{1}{9} \mathrm{dx}}{x - 1} + \int \frac{\frac{5}{27} \mathrm{dx}}{x - 4}$

$= - \frac{2}{27} \ln \left(| x + 5 |\right) - \frac{1}{9} \ln \left(| x - 1 |\right) + \frac{5}{27} \ln \left(| x - 4 |\right) + C$