How do you find int (x^2 - 4) / (x -1)^3 dxx24(x1)3dx using partial fractions?

1 Answer
Dec 27, 2017

The answer is =3/(2(x-1)^2)-2/(x-1)+ln(|x-1|)+C=32(x1)22x1+ln(|x1|)+C

Explanation:

Perform the decomposition into partial fractions

(x^2-4)/(x-1)^3=A/(x-1)^3+B/(x-1)^2+C/(x-1)x24(x1)3=A(x1)3+B(x1)2+Cx1

=(A+B(x-1)+C(x-1)^2)/((x-1)^3)=A+B(x1)+C(x1)2(x1)3

The denominators are the same, compare the numerators

x^2-4=A+B(x-1)+C(x-1)^2x24=A+B(x1)+C(x1)2

Let x=1x=1, =>, -3=A3=A

Coefficient of x^2x2

1=C1=C

Coefficients of xx

0=B-2C0=B2C, =>, B=2C=2B=2C=2

So,

(x^2-4)/(x-1)^3=-3/(x-1)^3+2/(x-1)^2+1/(x-1)x24(x1)3=3(x1)3+2(x1)2+1x1

int((x^2-4)dx)/(x-1)^3=-3intdx/(x-1)^3+2intdx/(x-1)^2+intdx/(x-1)(x24)dx(x1)3=3dx(x1)3+2dx(x1)2+dxx1

=3/(2(x-1)^2)-2/(x-1)+ln(|x-1|)+C=32(x1)22x1+ln(|x1|)+C