# How do you integrate [6/(x-3)^4*(x^2-4x+4)] using partial fractions?

##### 2 Answers

See the answer below:

Dec 22, 2016

$\frac{- 6 {x}^{2} + 30 x - 38}{x - 3} ^ 3 + C$

#### Explanation:

You don't. Instead, substitute $x = u + 3$, $\mathrm{dx} = \mathrm{du}$ and the integral becomes $6 \int \frac{1}{u} ^ 2 + \frac{3}{u} ^ 3 + \frac{1}{u} ^ 4 \mathrm{du}$
$= 6 \left(- \frac{1}{u} - \frac{1}{u} ^ 2 - \frac{\frac{1}{3}}{u} ^ 3\right) + C$ leading, upon substituting $u = x - 3$, to the answer. This substitution always works well if the denominator is simply a power of a linear expression.