Question

Evaluate the integral for x > 1/2:

integral x^2/sqrt(36x^2-9)

Answer 1

Let
\[x = \frac{\sec u}{2}\]
\[dx = \frac{\sec u \tan u}{2}\]
Note
sec^2 -1 = tan^2
\[\rightarrow \frac{1}{24} \int\limits \frac{ \sec^{2} u}{\tan u} (\sec u \tan u) du\]

Answer 2

is that the answer or do I have to take the integral of that?

Answer 3

no sorry i didnt finish, i just got you started by doing the substitution you still need to integrate