How do you integrate x^3/ (x^2 -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
=1/2(x^2 + ln(abs(x-1)) + ln(abs(x+1))) + C
Explanation:
x^3/(x^2-1)
= (x^3-x+x)/(x^2-1)
= (x(x^2-1)+x)/(x^2-1)
= x + x/(x^2-1)
= x + x/((x-1)(x+1))
= x + A/(x-1) + B/(x+1)
= x + (A(x+1)+B(x-1))/(x^2-1)
= x + ((A+B)x + (A-B))/(x^2-1)
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
