How do you use partial fraction decomposition to decompose the fraction to integrate (4x-2)/[3(x-1)^2]?

Jun 15, 2015

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

Explanation:

$\frac{4 x - 2}{3 {\left(x - 1\right)}^{2}} = \frac{1}{3} \cdot \frac{4 x - 2}{x - 1} ^ 2$

So we really only need to decompose the second factor:

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

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

$= \frac{A x - A + B}{x - 1} ^ 2$

So we need:

$A x - A + B = 4 x - 2$.

So $A = 4$ and, since $- A + B = - 2$, we get $B = 2$

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

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