# How do you integrate int(x+1)/((x+5)(x+2)(x-5)) using partial fractions?

Jan 10, 2017

The answer is =-2/15(∣x+5∣)+1/21ln(∣x+2∣)+3/35ln(∣x-5∣)+C

#### Explanation:

Perform the decomposition into partial fractions

$\frac{x + 1}{\left(x + 5\right) \left(x + 2\right) \left(x - 5\right)} = \frac{A}{x + 5} + \frac{B}{x + 2} + \frac{C}{x - 5}$

$= \frac{A \left(x + 2\right) \left(x - 5\right) + B \left(x + 5\right) \left(x - 5\right) + C \left(x + 5\right) \left(x + 2\right)}{\left(x + 5\right) \left(x + 2\right) \left(x - 5\right)}$

Therefore, by equating the numerators

$\left(x + 1\right) = \left(A \left(x + 2\right) \left(x - 5\right) + B \left(x + 5\right) \left(x - 5\right) + C \left(x + 5\right) \left(x + 2\right)\right)$

Let $x = - 5$, $\implies$, $- 4 = 30 A$, $\implies$, $A = - \frac{2}{15}$

Let $x = - 2$, $\implies$, $- 1 = - 21 B$, $\implies$, $B = \frac{1}{21}$

Let $x = 5$, $\implies$, $6 = 70 C$, $\implies$, $C = \frac{3}{35}$

so,

$\frac{x + 1}{\left(x + 5\right) \left(x + 2\right) \left(x - 5\right)} = \frac{- \frac{2}{15}}{x + 5} + \frac{\frac{1}{21}}{x + 2} + \frac{\frac{3}{35}}{x - 5}$

Now, perform the integration

$\int \frac{\left(x + 1\right) \mathrm{dx}}{\left(x + 5\right) \left(x + 2\right) \left(x - 5\right)} = - \frac{2}{15} \int \frac{\mathrm{dx}}{x + 5} + \frac{1}{21} \int \frac{\mathrm{dx}}{x + 2} + \frac{3}{35} \int \frac{\mathrm{dx}}{x - 5}$

=-2/15(∣x+5∣)+1/21ln(∣x+2∣)+3/35ln(∣x-5∣)+C