How do you find #int (3x^2 - 10) / (x^2-4x+4) dx# using partial fractions?

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Answer 1 · 2016-02-27T17:17:04

#3x+12lnabs(x-2)-2/(x-2)+C#

First, since the degrees of the numerator and denominator are equal, use polynomial long division to rewrite the expression:

#(3x^2-10)/(x^2-4x+4)=3+(12x-22)/(x^2-4x+4)#

Now, perform partial fraction decomposition on #(12x-22)/(x^2-4x+4)#, recognizing that #x^2-4x+4=(x-2)^2#.

#(12x-22)/((x-2)^2)=A/(x-2)+B/(x-2)^2#

Note that since the term is squared, it will be repeated.

Multiply both sides by #(x-2)^2# to see that

#12x-22=A(x-2)+B#

When we set #x=2#, we see that

#12(2)-22=A(0)+B#

#2=B#

Arbitrarily, set #x=3# to solve for #A#, recalling that #B=2#:

#12(3)-22=A(1)+2#

#A=12#

Thus,

#(3x^2-10)/(x^2-4x+4)=3+12/(x-2)+2/(x-2)^2#

Now, we can integrate more simply:

#int3+12/(x-2)+2/(x-2)^2dx#

#=3x+12lnabs(x-2)-2/(x-2)+C#