what is the integral of (sqr(x^2-16))/(x^4) dx?

Answer 1

Use the substitution x = 4sec(t). Then dx=4sec(t)tan(t) dt, and sqrt(x^2-16)=4tan(t). You're left with the integral of

tan^2(t) / 16sec^3(t) dt = (1/16) sin^2(t)cos(t) dt

which you can evaluate using another substitution, u=sin t.

You'll need to undo all of your substitutions to get to the final answer, keeping in mind that expressions of the form sec(t) can be replaced by x/4, and tan(t) can be replaced by sqrt(x^2-16)/4.

Answer 2

∫√(x^2-16)) / (x^4) dx = (x^2 -16)^(3/2)) / (48x^3) + c

Good Luck Mac

Answer 3

Homework: let x=4sec(theta).