How do you integrate x^3/ (x^2 -1)x3x2−1 using partial fractions?
1 Answer
Jun 9, 2016
int x^3/(x^2-1) dx = int (x + 1/(2(x-1)) + 1/(2(x+1))) dx∫x3x2−1dx=∫(x+12(x−1)+12(x+1))dx
=1/2(x^2 + ln(abs(x-1)) + ln(abs(x+1))) + C=12(x2+ln(|x−1|)+ln(|x+1|))+C
Explanation:
x^3/(x^2-1)x3x2−1
= (x^3-x+x)/(x^2-1)=x3−x+xx2−1
= (x(x^2-1)+x)/(x^2-1)=x(x2−1)+xx2−1
= x + x/(x^2-1)=x+xx2−1
= x + x/((x-1)(x+1))=x+x(x−1)(x+1)
= x + A/(x-1) + B/(x+1)=x+Ax−1+Bx+1
= x + (A(x+1)+B(x-1))/(x^2-1)=x+A(x+1)+B(x−1)x2−1
= x + ((A+B)x + (A-B))/(x^2-1)=x+(A+B)x+(A−B)x2−1
Equating coefficients we find:
{ (A+B = 1), (A-B = 0) :}
Hence:
A = B = 1/2
So:
x^3/(x^2-1) = x + 1/(2(x-1)) + 1/(2(x+1))
So:
int x^3/(x^2-1) dx = int (x + 1/(2(x-1)) + 1/(2(x+1))) dx
=1/2(x^2 + ln(abs(x-1)) + ln(abs(x+1))) + C
