Find the exact value of $\int^\infty_0\frac{5x^{2/3}}{-2-3x^{12/5}} dx$ Find the exact value of $$\int^\infty_0\frac{5x^\frac{2}{3}}{-2-3x^\frac{12}{5}} dx$$
I've tried using integration by parts. If you make $U=x^{10/15}$ then you end up with a more complicated expression.
If you make $u=-2-3x^\frac{12}{5}$ then we end up with an expression that is of equal difficulty. 
Subsequently I think neither of these directions are the way to go and would appreciate a nudge in the right direction. If someone thinks that either of those $u$'s actually would work than I'd be happy to write the work I did.
If $U=x^\frac{1}{15} \text{ then }du=\frac{1}{15x^\frac{14}{15}}$ already this seems very unwieldy and I'm not sure how to fit in.
 A: Hint: Factor $-\dfrac52$ forcefully outside of the integral sign, then let $t^\tfrac{12}5=\dfrac32~x^\tfrac{12}5$ and 
$u=\dfrac1{1+t^\tfrac{12}5}~,~$ and recognize the expression of the beta function in the new integral, 
then use Euler's reflection formula for the $\Gamma$ function to simplify the expression.
A: In math there are all sorts of powerful tools to evaluate definite integrals of this sort. You can use contour integration to tackle this integral.
First, we make the substitution $x=u^{15}$. This turns the integral to 
$$\int_{0}^{\infty}\dfrac{75u^{24}}{-2-3u^{36}} du$$
Now we do complex analysis. Promote $u$ to a complex variable; we are now integrating over the half-real line in the complex plane. But we can do better: take a path $C$ in the complex plane: $C_1+C_2+C_3$. $C_1$ is the real line; the part we want for our solution. $C_2$ is a path at $r=\infty$, and $C_3$ is a radial path leading back to the origin; take it to be at an angle $\pi/18$. Then use Cauchy's Residue Theorem.
$$\oint_{C}\dfrac{75u^{24}}{-2-3u^{36}} du = 2\pi i Res(C)$$
I'll leave the rest to you.
