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,

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

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

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

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