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

Feb 29, 2016

$\int \frac{x}{x - 3} ^ 2 \mathrm{dx} = \ln \left(| x - 3 |\right) - \frac{3}{x - 3} + C$, where $C$ is the constant of integration.

#### Explanation:

Using the partial fractions decomposition, we write $\frac{x}{x - 3} ^ 2$ in the form of $\frac{A}{x - 3} + \frac{B}{x - 3} ^ 2$, which is easier to integrate.

$\frac{x}{x - 3} ^ 2 = \frac{A}{x - 3} + \frac{B}{x - 3} ^ 2$

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

$\iff A = 1 \mathmr{and} - 3 A + B = 0 \iff A = 1 \mathmr{and} B = 3.$

Therefore, $\frac{x}{x - 3} ^ 2 = \frac{1}{x - 3} + \frac{3}{x - 3} ^ 2.$

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

$\int \frac{x}{x - 3} ^ 2 \mathrm{dx} = \ln \left(| x - 3 |\right) - \frac{3}{x - 3} + C$, where $C$ is the constant of integration.