Calculate the following integral:

Question

∫ (square root(9x^2 - 16)) / x^2

I found:

∫ (square root((3x)^2 - (4)^2)) / x^2

x = a sec0
3x = 4 sec0
x = 4/3 sec0
dx = 4/3 sec0tan0 d0

4/3 ∫ [(square root(4sec0)^2 - 16)(sec0tan0 d0)] / x^2

I'm lost from here and don't know how to simplify.

Answer 1

∫√(9x^2 - 16) / x^2 dx

let 3x = 4secθ
so,

9x^2 = 16sec^2 θ
9x^2 - 16 = 16sec^2 θ - 16 = 16tan^2 θ
√(9x^2 - 16) = 4tanθ
3 dx = 4secθtanθ dθ
x^2 = 16/9 sec^2 θ

∫√(9x^2 - 16) / x^2 dx
= ∫4tanθ / (16/9 sec^2 θ) * (4/3 secθtanθ) dθ
= ∫3tan^2 θ secθ dθ

Use integration by parts on that. Double check my math here.

Answer 2

Thank you Steve