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

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Answer 1 · 2017-12-27T07:29:36

The answer is $=3/(2(x-1)^2)-2/(x-1)+ln(|x-1|)+C$

Perform the decomposition into partial fractions

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

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

The denominators are the same, compare the numerators

$x^2-4=A+B(x-1)+C(x-1)^2$

Let $x=1$ , $=>$ , $-3=A$

Coefficient of $x^2$

$1=C$

Coefficients of $x$

$0=B-2C$ , $=>$ , $B=2C=2$

So,

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

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

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

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