How do you evaluate the integral #int (x^3-4)/(x+1)#?
1 Answer
Jan 22, 2017
#int(x^3-4)/(x+1)dx#
Let
#=int((u-1)^3-4)/udu#
Expand
#=int((u^3-3u^2+3u-1)-4)/u#
#=int(u^2-3u+3-5/u)du#
Integrate term by term:
#=1/3u^3-3/2u^2+3u-5lnabsu+C#
#=1/3(x+1)^3-3/2(x+1)^2+3(x+1)-5lnabs(x+1)+C#