Which is the right trig substitution to solve this integral?

Question

Which is the right trig substitution to solve this integral? $$\int \frac{dx}{(x^2-6x+3)^{\frac{3}{2}}} $$

Answer 1

The denominator is $((x-3)^2-6)^{3/2}$. So the most natural thing to try is $x-3=\sqrt{6}\sec(\theta)$.

Indeed this works, and simplies things - I am sure you can do the simplifications and get an answer easily.

Answer 2

$$\dfrac{1}{(x^2 - 6x + 3)^{3/2}} = \dfrac 1{[(x-3)^2 - 6]^{3/2}}$$

Let $(x - 3) = \sqrt 6\sec\theta \implies dx = \sqrt 6 \sec \theta \tan\theta\,d\theta$.

Answer 3

I think I would first complete to square and will get $((x-3)^2-6)^{3/2}$ and after that will substitute $x-3=\sqrt{6}\sec(u)$