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#