Find function defined in integral $\int_{0}^{1} \frac{4t\sqrt{t(1-t)}}{x-4t}dt $

Question

How can the function $$f(x) := \int_{0}^{1} \frac{4t\sqrt{t(1-t)}}{x-4t}dt $$

be calculated for $x>4$?

This integral came out when I'm trying to calculate $\sum_{k=1}^{N} \frac{C_{k}}{N^{k}}$ in an integral form. ($C_{n}$ is the $n$th Catalan number)

Answer 1

Let $t=\cos^2 y$. Then \begin{align} &\int_{0}^{1} \frac{4t\sqrt{t(1-t)}}{x-4t}dt =\int_0^{\pi/2}\frac{4\cos^4y\sin^2y}{x-4\cos^2y}dy.\\ =& \int_0^{\pi/2} \left(\frac{x^2}8 - \frac{x^2}8 \frac{x-4}{x-4\cos^2y} -\frac x2\sin^2t -\frac12\sin^22t \right) dy\\ =& \frac\pi{16}\left( {x^2} - x\sqrt{x(x-4)}-2x-2\right) \end{align}