<s>[INST] how do I evaluate this definite integral? $I=\int_3^5{\left( \frac{x^2+x}{\sqrt[3]{(x-3)(5-x)}} \right)dx}$ $$I=\int_3^5{\left( \frac{x^2+x}{\sqrt[3]{(x-3)(5-x)}} \right)dx}$$
I tried using substitution and some common definite integration results like
$$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx \\
 \int_a^bf(x)dx = (b-a)\int_0^1f((b-a)x+a)dx$$
and done substitutions like
$$x=3\cos^2(t)+5\sin^2(t)$$
but it just becomes messy.
I have tried adding the different results hoping that stuff would cancel out or simplify,
but I'm not able to figure it out.
Please help. Thanks!
 [/INST] Let $x=2t+3$, then $dx=2dt$. The bounds of the integral change like this: $x=3=2t+3\implies t=0$ and $x=2t+3=5\implies t=1$. Hence,
$\begin{align}
\int_3^5\frac{x^2+x}{\sqrt[3]{(x-3)(5-x)}}&=\int_0^1\frac{4t^2+14t+12}{\sqrt[3]{2t(2-2t)}}2dt\\
&=\sqrt[3]{2}\int_0^1\frac{4t^2+14t+12}{\sqrt[3]{t(1-t)}}dt\\
&=\sqrt[3]{2}\int_0^1 (4t^2+14t+12)t^{-\frac{1}{3}}(1-t)^{-\frac{1}{3}}dt\\
&=4\sqrt[3]{2}\int_0^1 t^{\frac{5}{3}}(1-t)^{-\frac{1}{3}}dt+
14\sqrt[3]{2}\int_0^1 t^{\frac{2}{3}}(1-t)^{-\frac{1}{3}}dt+
12\sqrt[3]{2}\int_0^1 t^{-\frac{1}{3}}(1-t)^{-\frac{1}{3}}dt\\
&=4\sqrt[3]{2}B(\frac{8}{3},\frac{2}{3})
+14\sqrt[3]{2}B(\frac{5}{3},\frac{2}{3})
+12\sqrt[3]{2}B(\frac{2}{3},\frac{2}{3})
\end{align}$
</s>