How do you integrate #(x+2)/(x^2+8x+16)# using partial fractions?

1 Answer
Dec 25, 2016

#int (x+2)/(x^2+8x+16) dx= ln|x+4|+2/(x+4)+C#

Explanation:

First you factorize the denominator of the rational function:

#x^2+8x+16 = (x+4)^2#

Now you can write:

#(x+2)/(x^2+8x+16) = (x+2)/((x+4)^2)= (x+4-2)/((x+4)^2)=1/(x+4)-2/((x+4)^2)#

The integral is then:

#int (x+2)/(x^2+8x+16) dx= int (dx)/(x+4)-2int(dx)/((x+4)^2)=ln|x+4|+2/(x+4)+C#