How do you evaluate the integral #int (x^3-4)/(x+1)#?

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Answer 1 · 2017-01-22T16:47:03

#int(x^3-4)/(x+1)dx#

Let #u=x+1#, implying that #du=dx#. This also means that #x=u-1#. Then:

#=int((u-1)^3-4)/udu#

Expand #(u-1)^3#:

#=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#