How do you integrate $int (x^2 - 5) / (x^2-4x+4)dx$ using partial fractions?

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Answer 1 · 2017-01-23T07:24:22

$int(x^2-5)/(x^2-4x+4)dx=x+4ln|x-2|+1/(x-2)+c$

$(x^2-5)/(x^2-4x+4)=(x^2-4x+4+4x-9)/(x^2-4x+4)$

= $1+(4x-9)/(x-2)^2$

Let $(4x-9)/(x-2)^2hArrA/(x-2)+B/(x-2)^2$

= $(A(x-2)+B)/(x-2)^2=(Ax+(B-2A))/(x-2)^2$

Therefore $A=4$ and $B-2A=-9$ and $B=-9+8=-1$

i.e. $(x^2-5)/(x^2-4x+4)=1+4/(x-2)-1/(x-2)^2$

and $int(x^2-5)/(x^2-4x+4)dx=int[1+4/(x-2)-1/(x-2)^2]dx$

= $x+4ln|x-2|+1/(x-2)+c$

How do you integrate int (x^2 - 5) / (x^2-4x+4)dxint (x^2 - 5) / (x^2-4x+4)dx using partial fractions?