# How do you find int (3x^2-x)/((1 - x)^2(1 - 3x))dx using partial fractions?

Nov 26, 2015

$\left[\ln \left(\frac{1}{x - 1}\right) + \frac{1}{x - 1}\right] + C$

#### Explanation:

$\int \frac{3 {x}^{2} - x}{{\left(1 - x\right)}^{2} \left(1 - 3 x\right)} \mathrm{dx}$

$\exists A , \exists B , \exists C$ such as

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

(I factorized the denominator to do partial fraction)

Multiply both side by : ${\left(x - 1\right)}^{2} \left(3 x - 1\right)$

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

$3 {x}^{2} - x = A \left(1 - 4 x + 3 {x}^{2}\right) + B \left(3 x - 1\right) + C \left(1 + {x}^{2} - 2 x\right)$

$3 {x}^{2} - x = A - 4 A x + 3 A {x}^{2} - B + 3 B x + C + C {x}^{2} - 2 C x$

$3 {x}^{2} - x = {x}^{2} \left(C + 3 A\right) + x \left(- 4 A + 3 B - 2 C\right) + A - B + C$

by identification

$C + 3 A = 3$
$- 4 A + 3 B - 2 C = - 1$
$A - B + C = 0$

you find ( ;) )

$A = 1$
$B = 1$
$C = 0$

and then

$\int \frac{3 {x}^{2} - x}{{\left(x - 1\right)}^{2} \left(3 x - 1\right)} \mathrm{dx} = - \int \frac{1}{x - 1} + \frac{1}{x - 1} ^ 2 \mathrm{dx}$

Which is $\left[\ln \left(\frac{1}{x - 1}\right) + \frac{1}{x - 1}\right] + C$

(Because $a \ln \left(b\right) = \ln \left({b}^{a}\right)$)

and $- \int \frac{1}{u} ^ 2 = \frac{1}{u}$