Indented Path Integration The goal is to show that 
$$\int_0^\infty \frac{x^{1/3}\log(x)}{x^2 + 1}dx = \frac{\pi^2}{6}$$
and that
$$\int_0^\infty \frac{x^{1/3}}{x^2 + 1}dx = \frac{\pi}{\sqrt{3}}.$$

So, we start with the function
$$f(z) = \frac{z^{1/3}\log(z)}{z^2 + 1}.$$

Let $c_r$ be the upper semicircle with radius r such that $r&lt; 1$ and let $c_R$ be the upper semicircle with radius R such that $R&gt;1$. Let $L_1$ be the interval $(r,R)$ on the reals and let $L_2$ be the interval $(-R,r)$ on the reals. 

Then
$$\int_{c_r} f + \int_{c_R} f + \int_{L_1} f + \int_{L_2} f$$
taken counterclockwise is equal to
$$2\pi i\text{ res}(f,i) = -\frac{\pi^2  e^{\frac{\pi i}{6}}}{2}$$

The integrals along $c_r$ and $c_R$ should go to $0$ as $r$ goes to $0$ and $R$ goes to infinity. 

So:
$$\int_{L_1} + \int_{L_2}= \int_0^{\infty}\frac{x^{1/3}\log(x)}{x^2 + 1}dx - \int_{0}^{\infty} \frac{x^{1/3}e^{i\pi/3}(ln(x) + i\pi) }{x^2 + 1}dx$$
where x is the polar radius. 

Let $A =  \int_0^\infty \frac{x^{1/3}\log(x)}{x^2 + 1}dx$
and let $B = \int_0^\infty \frac{x^{1/3}}{x^2 + 1}dx $.

Then we have $A*(1-e^{\frac{i\pi}{3}}) - \pi i e^{\frac{i\pi}{3}}B =  $-$\frac{\pi^2  e^{\frac{\pi i}{6}}}{2}$.

This doesn't mesh with the given values for $A$ and $B$, so I'm wondering where I went wrong. Thanks in advance.
In this case, I would use a keyhole contour about the positive real axis.  Consider the contour integral

$$\oint_C dz \frac{z^{1/3}}{1+z^2} \log{z}$$

where $C$ is the keyhole contour.  As it is clear that the integral about the circular arcs, large and small, vanish as their respective radii go to infinity and zero, the contour integral is equal to

$$\int_0^{\infty} dx \frac{x^{1/3}}{1+x^2} \log{x} - e^{i 2 \pi/3} \int_0^{\infty} dx \frac{x^{1/3}}{1+x^2} (\log{x}+i 2 \pi) $$

which is equal to

$$\left (1-e^{i 2 \pi/3} \right )\int_0^{\infty} dx \frac{x^{1/3}}{1+x^2} \log{x} - i 2 \pi e^{i 2 \pi/3} \int_0^{\infty} dx \frac{x^{1/3}}{1+x^2}$$

By the residue theorem, the contour integral is $i 2 \pi$ times the sum of the residues at the poles $z=e^{i \pi/2}$ and $z=e^{i 3 \pi/2}$, or

$$i 2 \pi \frac{e^{i \pi/6}}{2 i} i \frac{\pi}{2} + i 2 \pi \frac{e^{i \pi/2}}{-2 i} i \frac{3 \pi}{2} = \frac{5 \pi^2}{4} + i \frac{\sqrt{3} \pi^2}{4}$$

As you did above, let $A$ be the first integral and $B$ be the second.  Then

$$\left (1-e^{i 2 \pi/3} \right ) A - i 2 \pi e^{i 2 \pi/3} B = \frac{5 \pi^2}{4} + i \frac{\sqrt{3} \pi^2}{4}$$

So we equate real and imaginary parts.  Rearranging, we get two equations and two unknowns:

$$\frac{3}{2} A + \sqrt{3} \pi B = \frac{5 \pi^2}{4} $$
$$-\frac{\sqrt{3}}{2} A + \pi B = \frac{\sqrt{3} \pi^2}{4}$$

From this, you may easily verify that

$$A = \int_0^{\infty}dx \frac{x^{1/3}}{1+x^2} \log{x} = \frac{\pi^2}{6}$$

$$B = \int_0^{\infty}dx \frac{x^{1/3}}{1+x^2} = \frac{\pi}{\sqrt{3}}$$
Indented contours pick up half a residue. So the integral along $c_r$ is not zero.