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

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Answer 1 · 2018-03-06T08:14:20

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

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

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

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

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

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

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

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

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

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