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

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Answer 1 · 2017-01-12T08:00:02

The answer is $=ln(∣x-1∣)-1/(x-1)-1/(2(x-1)^2)+C$

We use
$intx^ndx=x^(n+1)/(n+1)+C (n!=-1)$

and $intdx/x=lnx +C$

We use the method by substitution

Let $u=x-1$ , $=>$ , $du=dx$

$x=u+1$

So,

$x^2-x+1=(u+1)^2-(u)$

$=u^2+2u+1-u$

$=u^2+u+1$

Therefore,

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

$=int((u^2+u+1)du)/(u^3)$

$=int(1/u+1/u^2+1/u^3)du$

$=lnu-1/u-1/(2u^2)$

$=ln(∣x-1∣)-1/(x-1)-1/(2(x-1)^2)+C$

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