# How do you integrate int (4x-2) /( 3(x-1)^2) using partial fractions?

May 12, 2016

$\int \frac{4 x - 2}{3 {\left(x - 1\right)}^{2}} \mathrm{dx} = \frac{4}{3} \ln \left(x - 1\right) - \frac{4}{3 \left(x - 1\right)} + c$

#### Explanation:

Let us first find partial fractions of $\frac{4 x - 2}{3 {\left(x - 1\right)}^{2}}$ and for this let

$\frac{4 x - 2}{{\left(x - 1\right)}^{2}} \Leftrightarrow \frac{A}{x - 1} + \frac{B}{x - 1} ^ 2$ or

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

Hence $A = 4$ and $B - A = - 2$ i.e. $B = 4 - 2 = 2$

Hence $\int \frac{4 x - 2}{3 {\left(x - 1\right)}^{2}} \mathrm{dx} = \frac{1}{3} \int \left[\frac{2}{x - 1} + \frac{2}{x - 1} ^ 2\right] \mathrm{dx}$

= $\frac{2}{3} \int \frac{2}{x - 1} \mathrm{dx} + \frac{2}{3} \int \frac{2}{x - 1} ^ 2 \mathrm{dx} + k$

= $\frac{4}{3} \ln \left(x - 1\right) - \frac{4}{3 \left(x - 1\right)} + c$