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

Dec 27, 2017

The answer is $= \frac{3}{2 {\left(x - 1\right)}^{2}} - \frac{2}{x - 1} + \ln \left(| x - 1 |\right) + C$

#### Explanation:

Perform the decomposition into partial fractions

$\frac{{x}^{2} - 4}{x - 1} ^ 3 = \frac{A}{x - 1} ^ 3 + \frac{B}{x - 1} ^ 2 + \frac{C}{x - 1}$

$= \frac{A + B \left(x - 1\right) + C {\left(x - 1\right)}^{2}}{{\left(x - 1\right)}^{3}}$

The denominators are the same, compare the numerators

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

Let $x = 1$, $\implies$, $- 3 = A$

Coefficient of ${x}^{2}$

$1 = C$

Coefficients of $x$

$0 = B - 2 C$, $\implies$, $B = 2 C = 2$

So,

$\frac{{x}^{2} - 4}{x - 1} ^ 3 = - \frac{3}{x - 1} ^ 3 + \frac{2}{x - 1} ^ 2 + \frac{1}{x - 1}$

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

$= \frac{3}{2 {\left(x - 1\right)}^{2}} - \frac{2}{x - 1} + \ln \left(| x - 1 |\right) + C$