Integration Problem?

Question

I'd appreciate if someone could help me understand how to integrate the following:
-1/((x^3)/(x^2-16)^(1/2))

Answer 1

∫-1/(x³/√(x²-16)) dx

First, simplify the fraction:

∫-√(x²-16)/x³ dx

Now, make the substitution x = 4 sec θ, so dx = 4 sec θ tan θ dθ. Thus we have:

∫-√(16 sec² θ-16)/(64 sec³ θ) * 4 sec θ tan θ dθ

Simplifying:

∫-√(16 sec² θ-16)/(16 sec² θ) * tan θ dθ
∫-√(16 tan² θ)/(16 sec² θ) * tan θ dθ
∫-tan² θ/(4 sec² θ) dθ
-1/4 ∫sin² θ dθ
-1/4 ∫1 - cos (2θ) dθ
-θ/4 + sin (2θ)/8 + C

Now, we just need to resubstitute x. First note that θ = arcsec (x/4), so we just need to find sin (2θ). Let y=sin (2θ). Then we have:

y=sin (2θ)
y=2 sin θ cos θ
y=2 √(1-cos² θ) cos θ
y=8 √(1-16/(16 sec² θ)) / (4 sec θ)
y=8 √(1-16/x²)/x

So substituting this into the original expression, we have:

-arcsec (x/4)/4 + √(1-16/x²)/x + C

And so we are done.