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#
