Improper integral $\int \limits_{2}^{4}\frac{\sqrt{(16 - x^2)^5}}{(x^2 - 9x + 20)^3}dx$ I can't figure out how to solve (say whether it converges or diverges) the following improper integral:
$$
\int \limits_{2}^{4}\frac{\sqrt{(16 - x^2)^5}}{(x^2 - 9x + 20)^3}dx
$$
I've tried to simplify this and got:
$$
\int \limits_{2}^{4}\frac{(4 + x)^{5/2}}{(4 - x)^{1/2}(5 - x)^3}
$$
But I don't know what to do next. Could you please give me some hint

An alternate substitution is $x = 4 \, \sin(t)$ which leads to, with some difficulty,
\begin{align}
\int \frac{(16 - x^2)^{5/2}}{(x^2 - 9 x + 20)^3} \, dx = \frac{81}{2} \, \frac{x-6}{(x-5)^2} \, \sqrt{16 - x^2} - 63 \, \tan^{-1}\left(\frac{16 - 5x}{3 \, \sqrt{16 - x^2}}\right) - \sin^{-1}\left(\frac{x}{4}\right) + c_{0}. 
\end{align}
Evaluating the integral at the limits (2,4) then provides the integrals value:
\begin{align}
\int_{2}^{4} \frac{(16 - x^2)^{5/2}}{(x^2 - 9 x + 20)^3} \, dx = 36 \, \sqrt{3} + \frac{125 \, \pi}{3}.
\end{align}
