How do you evaluate the integral #int dx/(x^4-16)#?

Question

9491 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-06T08:14:20

# 1/32ln|(x-2)/(x+2)|-1/16arc tan(x/2)+C#.

Let, #I=intdx/(x^4-16)=intdx/{(x^2+4)(x^2-4)}#

#=1/8int8/{(x^2+4)(x^2-4)}dx#,

#=1/8int{(x^2+4)-(x^2-4)}/{(x^2+4)(x^2-4)}dx#,

#=1/8int{(x^2+4)/{(x^2+4)(x^2-4)}-(x^2-4)/{(x^2+4)(x^2-4)}}dx#,

#=1/8int{1/(x^2-4)-1/(x^2+4)}dx#,

#=1/8{int1/(x^2-2^2)-int1/(x^2+2^2)dx}#,

#=1/8{1/(2xx2)ln|(x-2)/(x+2)|-1/2arc tan(x/2)}#,

#rArr I=1/32ln|(x-2)/(x+2)|-1/16arc tan(x/2)+C#.