<s>[INST] Evaluate the improper integral: $\int _0^{+\infty }\:\frac{dx}{\left(x^2+9\right)\cdot \sqrt[3]{x^2+9}}$ $$\int _0^{+\infty }\:\frac{dx}{\left(x^2+9\right)\cdot \sqrt[3]{x^2+9}}$$
Wolfram says strange things. And you need to solve analytically. Help.
 [/INST] $$\begin{align*}I=\int_{0}^{+\infty}(x^2+9)^{-4/3}\,dx = 3^{-5/3}\int_{0}^{+\infty}(1+z^2)^{-4/3}\,dz=\frac{\sqrt{\pi}\cdot\Gamma\left(\frac{5}{6}\right)}{2\cdot 3^{5/3}\cdot\Gamma\left(\frac{4}{3}\right)}\end{align*}$$
by using the substitution $z=\tan\theta$ and Euler's beta function. That simplifies to:
$$ I =\frac{\pi^{3/2}}{3^{2/3}\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{1}{6}\right)}. $$
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