How do you evaluate the integral #int (x^3-1)/(x^3-x^2)#?

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Answer 1 · 2017-05-01T09:58:50

The answer is #=x+ln(|x|)-1/x+C#

We need

#a^3-b^3=(a-b)(a^2+ab+b^2)#

#intx^ndx=x^(n+1)/(n+1)+C(n!=-1)#

#intdx/x=ln(|x|)+C#

Let start by doing some simplification

#(x^3-1)/(x^3-x^2)=((x-1)(x^2+x+1))/(x^2(x-1))#

#=(x^2+x+1)/x^2#

#=1+1/x+1/x^2#

Therefore,

#int((x^3-1)dx)/(x^3-x^2)=int(1+1/x+1/x^2)dx#

#=int1*dx+intdx/x+intdx/x^2#

#=x+ln(|x|)-1/x+C#