# How do you integrate int (2x+1)/((x+8)(x-2)(x-3))  using partial fractions?

Jan 31, 2017

The answer is $= - \frac{3}{22} \ln \left(| x + 8 |\right) - \frac{1}{2} \ln \left(| x - 2 |\right) + \frac{7}{11} \ln \left(| x - 3 |\right) + C$

#### Explanation:

Let's perform the decomposition into partial fractions

$\frac{2 x + 1}{\left(x + 8\right) \left(x - 2\right) \left(x - 3\right)} = \frac{A}{x + 8} + \frac{B}{x - 2} + \frac{C}{x - 3}$

$= \frac{A \left(x - 2\right) \left(x - 3\right) + B \left(x + 8\right) \left(x - 3\right) + C \left(x + 8\right) \left(x - 2\right)}{\left(x + 8\right) \left(x - 2\right) \left(x - 3\right)}$

As the denominators are the same, we can equalise the numerators

$\left(2 x + 1\right) = A \left(x - 2\right) \left(x - 3\right) + B \left(x + 8\right) \left(x - 3\right) + C \left(x + 8\right) \left(x - 2\right)$

Let $x = - 8$, $\implies$, $- 15 = 110 A$, $\implies$, $A = - \frac{3}{22}$

Let $x = 2$, $\implies$, $5 = - 10 B$, $\implies$, $B = - \frac{1}{2}$

Let $x = 3$, $\implies$, $7 = 11 C$, $\implies$, $C = \frac{7}{11}$

Therefore,

$\frac{2 x + 1}{\left(x + 8\right) \left(x - 2\right) \left(x - 3\right)} = \frac{- \frac{3}{22}}{x + 8} + \frac{- \frac{1}{2}}{x - 2} + \frac{\frac{7}{11}}{x - 3}$

So,

$\int \frac{\left(2 x + 1\right) \mathrm{dx}}{\left(x + 8\right) \left(x - 2\right) \left(x - 3\right)} = - \frac{3}{22} \int \frac{\mathrm{dx}}{x + 8} - \frac{1}{2} \int \frac{\mathrm{dx}}{x - 2} + \frac{7}{11} \int \frac{\mathrm{dx}}{x - 3}$

$= - \frac{3}{22} \ln \left(| x + 8 |\right) - \frac{1}{2} \ln \left(| x - 2 |\right) + \frac{7}{11} \ln \left(| x - 3 |\right) + C$