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

Mar 3, 2016

$\frac{3}{7} \ln | x - 2 | - \frac{2}{21} \ln | x + 5 | - \frac{1}{3} \ln | x - 1 | + c$

#### Explanation:

Since the factors on the denominator are linear , the the numerators of the partial fractions will be constants , say A , B and C.

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

multiply through by (x+5)(x-1)(x-2)

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

The aim now is to find the values of A,B and C. Note that if x=1 then the terms with A and C will be zero. If x =2 the terms with A and B will be zero and if x = -5 the terms with B and C will be zero. This is the starting point in finding A , B and C.

let x = 1 in (1): 2 = -6B $\Rightarrow B = - \frac{1}{3}$

let x = 2 in (1): 3 = 7C$\Rightarrow C = \frac{3}{7}$

let x = -5 in (1): -4 = 42A$\Rightarrow A = - \frac{2}{21}$

Integral can now be written as :

$\int \frac{- \frac{2}{21}}{x + 5} \mathrm{dx} - \int \frac{\frac{1}{3}}{x - 1} \mathrm{dx} + \int \frac{\frac{3}{7}}{x - 2} \mathrm{dx}$

$= \frac{3}{7} \ln | x - 2 | - \frac{2}{21} \ln | x + 5 | - \frac{1}{3} \ln | x - 1 | + c$