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How do you integrate #int (x^2-2x+1)/(x-2)^3# using partial fractions?

Answer 1

Isaiah Allen

# ln|x-2|-2/(x-2)-1/{2(x-2)^2}+C.#

Without breaking down the problem, we can solve it.

As demonstrated below, integrateand into partial fractions

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

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

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

#=int[{(x-2)+2}/(x-2)^2+1/(x-2)^3]dx#

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

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

We now apply the subsequent result:

#intf(x)dx=F(x)+crArrintf(ax+b)=1/aF(ax+b),ane0#.

[ It can be easily proved by subst.ing, for #ax+b#.]

#:. I=ln|x-2|-2/(x-2)-1/{2(x-2)^2}+C.#

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Answer 2

Charles Anctil

# ln|x-2|-2/(x-2)-1/{2(x-2)^2}+C.#

Noticing that #(x-2)^3# is a perfect cube, there is no need for partial fractions. Substitute #u=x-2# and #(du)/(dx)=1# and the integral becomes: #int((u+2)^2-2(u+2)+1)/u^3 xx 1 du# #=int(u^2+2u+1)/u^3du# #=int1/u+2/u^2+1/u^3du# #= ln|x-2|-2/(x-2)-1/{2(x-2)^2}+C.#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.