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

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Answer 1 · 2015-08-30T16:13:40

First perform the division.

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

$x^4/((x-1)^3) = x^4/(x^3-3x^2+3x-1)$

$= x+ (3x^2-3x+1)/(x-1)^3$

To find the partial fraction decomposition of $(3x^2-3x+1)/(x-1)^3$ , find $A, B " and", C$ so that:

$A/(x-1)+B/(x-1)^2 + C/(x-1)^3 = (3x^2-3x+1)/(x-1)^3$

Clear the denominators to get:

$A(x^2-2x+1)+B(x-1)+C = 3x^2-3x+1$

So $A=3$ and

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

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

$x^4/((x-1)^3) = x+3/(x-1)-3/(x-1)^2 + 1/(x-1)^3$

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