# Determining k: $\int_{6}^{16} \frac{dx}{\sqrt{x^3 + 7x^2 + 8x – 16}} = \frac{\pi}{k}$

I have a calculus II final coming up and this question came up in a past final exam:

$$\int_{6}^{16} \frac{dx}{\sqrt{x^3 + 7x^2 + 8x – 16}} = \frac{\pi}{k},$$

where $k$ is a constant. Find $k$.

My progress so far:

$$\int_{6}^{16} \frac{dx}{\sqrt{(x – 1)}(x + 4)} = \frac{\pi}{k}$$

The answer is: $k = 6\sqrt{5}$

I do not know where to from from this step. Any helps or hints will be greatly appreciated! Thank you.

EDIT: $(x + 4)$ not $(x + 2)$ in the denominator.

#### Solutions Collecting From Web of "Determining k: $\int_{6}^{16} \frac{dx}{\sqrt{x^3 + 7x^2 + 8x – 16}} = \frac{\pi}{k}$"

Use substitution, put say $u = \sqrt{x-1}$ and then see what happens. It is useful to know that the derivative of $\tan^{-1} x$ is $\frac{1}{x^2+1}$.