<s>[INST] Simpler way to compute a definite integral without resorting to partial fractions? I found the method of partial fractions very laborious to solve this definite integral :
$$\int_0^\infty \frac{\sqrt[3]{x}}{1 + x^2}\,dx$$
Is there a simpler way to do this ?
 [/INST] Let's generalize the problem. We will evaluate
$$
\int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx.
$$
Let $$y=\dfrac{1}{1+x^b}\quad\Rightarrow\quad x=\left(\dfrac{1-y}{y}\right)^{\large\frac1b}\quad\Rightarrow\quad dx=-\left(\dfrac{1-y}{y}\right)^{\large\frac1b-1}\ \dfrac{dy}{by^2}\ ,$$ then
\begin{align}
\int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx&=\int_0^1 y\left(\dfrac{1-y}{y}\right)^{\large\frac{a-1}b}\left(\dfrac{1-y}{y}\right)^{\large\frac1b-1}\ \dfrac{dy}{by^2}\\&=\frac1b\int_0^1y^{\large1-\frac{a}{b}-1}(1-y)^{\large\frac{a}{b}-1}\ dy,
\end{align}
where the last integral in RHS is Beta function.
$$
\text{B}(x,y)=\int_0^1t^{\ \large x-1}\ (1-t)^{\ \large y-1}\ dt=\frac{\Gamma(x)\cdot\Gamma(y)}{\Gamma(x+y)}.
$$
Hence
\begin{align}
\int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx&=\frac1b\int_0^1y^{\large1-\frac{a}{b}-1}(1-y)^{\large\frac{a}{b}-1}\ dy\\&=\frac1b\cdot\Gamma\left(1-\frac{a}{b}\right)\cdot\Gamma\left(\frac{a}{b}\right)\\&=\large{\color{blue}{\frac{\pi}{b\sin\left(\frac{a\pi}{b}\right)}}}.
\end{align}
The last part uses Euler's reflection formula for Gamma function provided $0<a<b$. Thus
$$
\large\int_0^\infty\dfrac{\sqrt[3]{x}}{1+x^2}\ dx=\int_0^\infty\dfrac{x^{\large\frac43-1}}{1+x^2}\ dx=\frac{\pi}{2\sin\left(\frac{2\pi}{3}\right)}=\color{blue}{\frac\pi3\sqrt{3}}.
$$
</s>