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

Aug 30, 2015

First perform the division.

Explanation:

In order to use partial fraction decomposition we must have the degree of the numerator less than the degree of the denominator.

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

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

To find the partial fraction decomposition of $\frac{3 {x}^{2} - 3 x + 1}{x - 1} ^ 3$, find $A , B \text{ and} , C$ so that:

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

Clear the denominators to get:

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

So $A = 3$ and

$- 2 A + B = - 3$, so $B = 3$

finally, $A - B + C = 1$, so $C = 1$

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