Simpler way to compute a definite integral without resorting to partial fractions?

Perhaps this is simpler.

Make the substitution $\displaystyle x^{2/3} = t$. Giving us

$\displaystyle \frac{2 x^{1/3}}{3 x^{2/3}} dx = dt$, i.e $\displaystyle x^{1/3} dx = \frac{3}{2} t dt$

This gives us that the integral is

$$I = \frac{3}{2} \int_{0}^{\infty} \frac{t}{1 + t^3} \ \text{d}t$$

Now make the substitution $t = \frac{1}{z}$ to get

$$I = \frac{3}{2} \int_{0}^{\infty} \frac{1}{1 + t^3} \ \text{d}t$$

Add them up, cancel the $\displaystyle 1+t$, write the denominator ($\displaystyle t^2 - t + 1$) as $\displaystyle (t+a)^2 + b^2$ and get the answer.

By using techniques of complex analysis ($\text{Residue Theory}$) one can actually show that $$\int\limits_{0}^{\infty} \frac{x^{a-1}}{1+x^{b}} \ \text{dx} = \frac{\pi}{b \sin(\pi{a}/b)}, \qquad 0 < a <b$$

You can obtain the value of your $\text{Integral}$ by putting $a=\frac{4}{3}$ and $b=2$.

Set $$I = \int\limits_{0}^{\infty} \frac{x^{a-1}}{1+x^{b}} \ \text{dx}$$ and integrate $$f(z) = \frac{z^{a-1}}{1+z^{b}} = \frac{|z|^{a-1} \cdot e^{i(a-1)\text{arg}(z)}}{1+|z|^{b}e^{ib\text{arg}(z)}}$$

Simple pole at $z_{1} = e^{\pi{i}/b}$ and hence $$\text{Res} \Biggl[\frac{z^{a-1}}{1+z^{b}}, e^{\pi{i}/b}\Biggr] = \frac{z^{a-1}}{bz^{b-1}}\Biggl|_{z =e^{\pi i / b}} = -\frac{1}{b}e^{\pi i a/b}$$

Integrate along $\gamma_{1}$, and let $R \to \infty$ and let $ \epsilon \to 0^{+}$. This gives, \begin{align*} \int\limits_{\gamma_{1}} f(z) \ dz & = \int\limits_{\gamma_{1}} \frac{|z|^{a-1} \cdot e^{i(a-1)\text{arg}(z)}}{1+|z|^{b}e^{ib\text{arg}(z)}} \ dz \\ &= \int\limits_{\epsilon}^{R} \frac{x^{a-1}}{1+x^{b}} \to \int\limits_{0}^{\infty} \frac{x^{a-1}}{1+x^{b}} \ dx =I \end{align*}

Integrate along $\gamma_{2}$, and let $R \to \infty$. This gives $0 < a < b$ and $$\Biggl|\int\limits_{\gamma_{2}} f(z) dz \Biggr| \leq \frac{R^{a-1}}{R^{b}-1} \cdot \frac{2\pi R}{b} \sim \frac{2 \pi}{b R^{b-a}} \to 0$$

Integrate along $\gamma_{3}$ and let $R \to \infty$ and $\epsilon \to 0^{+}$. This gives \begin{align*} \int\limits_{\gamma_{3}} f(z) \ dz &= \int\limits_{\gamma_{3}} \frac{|z|^{a-1} \cdot e^{i(a-1)\text{arg}(z)}}{1+|z|^{b}e^{ib\text{arg}(z)}} = \Biggl[\begin{array}{c} z=x e^{2\pi i/b} \\ dz=e^{2\pi i/b} \ dx \end{array}\Biggr] \\ &= \int\limits_{R}^{\epsilon} \frac{x^{a-1}e^{2\pi i(a-1)/b}}{1+x^{b}} \cdot e^{2\pi i b} \ dx \to \int\limits_{\infty}^{0} \frac{x^{a-1}e^{2\pi i(a-1)/b}}{1+x^{b}} \cdot e^{2\pi i b} \ dx \\ &= -e^{2\pi ia/b}I \end{align*}

Integrate along $\gamma_{4}$ and let $\epsilon \to 0^{+}$. This gives $0 < a <b$, $$\Biggl|\int\limits_{\gamma_{4}} f(z) \ dz \Biggr| \leq \frac{\epsilon^{a-1}}{1-\epsilon^{b}} \cdot \frac{2\pi\epsilon}{b} \sim \frac{2\pi\epsilon}{b} \to 0$$

Using the $\text{Residue Theorem}$ and letting $R \to \infty$ and $\epsilon \to 0^{+}$, we obtain that $$ I + 0 - e^{2\pi a/b}I + 0 = 2\pi i \cdot \Bigl(-\frac{1}{b} e^{\pi ia/b}\Bigr)$$ This yields, $$(e^{-\pi i a/b} - e^{\pi i a./b})I= -\frac{2\pi i}{b}$$ and hence solving for $I$, we have $$I= \frac{2\pi i}{b \cdot (e^{\pi ia/b} - e^{-\pi i a/b})}=\frac{\pi}{b \sin(\pi a/b)}$$

Here is a different way I am quite fond of:

Call our integral I, that is set $$I=\int_0^\infty \frac{\sqrt[3]{x}}{1+x^2}\,dx$$ Let $u=1+x^{2}$ so that $du=2x \, dx$. Since $$\sqrt[3]{x} \, dx=\frac{1}{2}\frac{2x \, dx}{\sqrt[3]{x^{2}}}=\frac{1}{2}\frac{u}{\left({u-1}\right)^{\frac{1}{3}}}\,du$$ we have that$$I=\frac{1}{2}\int_{1}^{\infty}\frac{1}{u\left(u-1\right)^{\frac{1}{3}}}\,du.$$ Let $u=\frac{1}{v}$ so that this becomes $$\frac{1}{2}\int_{0}^{1}\frac{1}{\frac{1}{v}\left(\frac{1}{v}-1\right)^{\frac{1}{3}}}\frac{1}{v^{2}}\,dv=\frac{1}{2}\int_{0}^{1}v^{-\frac{2}{3}}\left(1-v\right)^{-\frac{1}{3}}\,dv=\frac{1}{2}\text{B}\left(\frac{1}{3},\frac{2}{3}\right)$$ where $\text{B}(x,y)$ is the beta function. Since $$\text{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ we have that $$I=\frac{1}{2}\frac{\Gamma(\frac{1}{3})\Gamma(\frac{2}{3})}{\Gamma(1)}.$$ Since $\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin\pi s}$ it follows that $$I=\frac{\pi}{2\sin\pi/3}=\frac{\pi}{\sqrt{3}}.$$ Hope that helps,

Also it is worth mentioning that numerically, Wolfram Alpha agrees with this answer.