# How do you integrate (x-5)/(x-2)^2 using partial fractions?

Dec 19, 2016

$\ln | x - 2 | + \frac{3}{x - 2} + C$

#### Explanation:

$\int \frac{x - 5}{x - 2} ^ 2 \mathrm{dx}$
$= \int \frac{x - 2 - 3}{x - 2} ^ 2 \mathrm{dx}$
$= \int \left(\frac{\cancel{x - 2}}{x - 2} ^ \cancel{2} - \frac{3}{x - 2} ^ 2\right) \mathrm{dx}$
$= \int \left(\frac{1}{x - 2} - \frac{3}{x - 2} ^ 2\right) \mathrm{dx}$
$= \ln | x - 2 | + \frac{3}{x - 2} + C$.

If the last step is not obvious, substitute $u = x - 2$. Alternatively, substitute $u = x - 2$ right at the start to get $\int \left(\frac{1}{u} - \frac{3}{u} ^ 2\right) \mathrm{du}$, but this isn't then really partial fractions. However, it isn't really a partial fractions question because you have only one unique term in the denominator.