TITLE: Having trouble evaluating this integral
QUESTION [5 upvotes]: $$\int\limits_6^{16}\left(\frac{1}{\sqrt{x^3+7x^2+8x-16}}\right)\,\mathrm dx=\frac{\pi }{k}$$
Note: $k$ is a constant.

REPLY [7 votes]: $$x^3+7x^2+8x-16=(x-1)(x+4)^2\implies \sqrt{x^3+7x^2+8x-16}=\sqrt{x-1}(x+4)$$
Denote the indefinite integral as $I$.
$$I=\int\frac{\mathrm dx}{(x+4)\sqrt{x-1}}$$
Make the substitution $u^2=x-1$ and $2u\,\mathrm du=\mathrm dx$ to get,
$$I=2\int\frac{\mathrm du}{u^2+5}$$
This is an elementary integral which can be evaluated easily. Put the limits and get the value of $k$