# How do you integrate int (2x+1)/((x-4)(x-1)(x+7))  using partial fractions?

Dec 9, 2016

The answer is =3/11ln(∣x-4∣)-1/8ln(∣x-1∣)-13/88ln(∣x+7∣)+C

#### Explanation:

Let's do the decomposition into partial fractions

$\frac{2 x + 1}{\left(x - 4\right) \left(x - 1\right) \left(x + 7\right)} = \frac{A}{x - 4} + \frac{B}{x - 1} + \frac{C}{x + 7}$

$= \frac{A \left(x - 1\right) \left(x + 7\right) + B \left(x - 4\right) \left(x + 7\right) + C \left(x - 1\right) \left(x - 4\right)}{\left(x - 4\right) \left(x - 1\right) \left(x + 7\right)}$

Therefore,

$2 x + 1 = A \left(x - 1\right) \left(x + 7\right) + B \left(x - 4\right) \left(x + 7\right) + C \left(x - 1\right) \left(x - 4\right)$

Let $x = 1$, $\implies$,$3 = - 24 B$, $\implies$, $B = - \frac{1}{8}$

Let $x = 4$, $\implies$,$9 = 33 A$, $\implies$, $A = \frac{3}{11}$

Let $x = - 7$, $\implies$, $- 13 = 88 C$, $\implies$$C = - \frac{13}{88}$

So,

$\frac{2 x + 1}{\left(x - 4\right) \left(x - 1\right) \left(x + 7\right)} = \frac{\frac{3}{11}}{x - 4} - \frac{\frac{1}{8}}{x - 1} - \frac{\frac{13}{88}}{x + 7}$

$\int \frac{\left(2 x + 1\right) \mathrm{dx}}{\left(x - 4\right) \left(x - 1\right) \left(x + 7\right)} = \frac{3}{11} \int \frac{\mathrm{dx}}{x - 4} - \frac{1}{8} \int \frac{\mathrm{dx}}{x - 1} - \frac{13}{88} \int \frac{\mathrm{dx}}{x + 7}$

=3/11ln(∣x-4∣)-1/8ln(∣x-1∣)-13/88ln(∣x+7∣)+C