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

Feb 18, 2016

$\frac{1}{4} \ln | 3 x - 2 | - \frac{1}{20} \ln | x + 2 | - \frac{1}{5} \ln | 2 x - 1 | + c$

#### Explanation:

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

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

Now multiply through by (3x-2)(x+2)(2x-1)

hence : x = A(x+2)(2x-1) + B(3x-2)(2x-1) + C(3x-2)(x+2)...(1)

The aim now is to find values of A , B and C. Note that if x =-2 , the terms with A and C will be zero.
If x$= \frac{1}{2} \text{ the terms with A and B will be zero}$
and if x$= \frac{2}{3} \text{the terms with B and C will be zero}$

let x = -2 in (1): - 2 = 40B $\Rightarrow B = - \frac{1}{20}$
let $x = \frac{1}{2} \text{ in (1)} : \frac{1}{2} = - \frac{5}{4} C \Rightarrow C = - \frac{2}{5}$
let $x = \frac{2}{3} \text{ in (1)} : \frac{2}{3} = \frac{8}{9} A \Rightarrow A = \frac{3}{4}$

hence integral becomes

$\int \frac{\frac{3}{4}}{3 x - 2} \mathrm{dx} - \frac{\frac{1}{20}}{x + 2} \mathrm{dx} - \frac{\frac{2}{5}}{2 x - 1} \mathrm{dx}$

$= \frac{3}{4} . \frac{1}{3} \ln | 3 x - 2 | - \frac{1}{20} \ln | x + 2 | - \frac{2}{5} . \frac{1}{2} \ln | 2 x - 1 | + c$

$= \frac{1}{4} \ln | 3 x - 2 | - \frac{1}{20} \ln | x + 2 | - \frac{1}{5} \ln | 2 x - 1 | + c$