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

Question

2513 views around the world

You can reuse this answer
Creative Commons License

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#