Evaluating the integral $\int_0^{\infty}\frac{dx}{\sqrt[4]{x}(1+x^2)}$ using Residue Theorem I need to evaluate the integral $$\int_0^{\infty}\frac{dx}{\sqrt[4]{x}(1+x^2)}$$
I've been given the hint to use the keyhole contour. So I would first choose the principal branch of $\sqrt[4]{\cdot}$, then I have the "keyhole" around $0$, giving me
$$2\pi i\left(\text{Res}_if+\text{Res}_{-i}f\right)= \\ \int_{\gamma_{R,\epsilon}}\frac{dz}{\sqrt[4]{z}(1+z^2)}=\int_{\gamma_R}f(z)dz-\int_{\gamma_\epsilon}f(z)dz+\int^R_{\epsilon} f(z)dz - \int^R_{\epsilon} f(-z)dz$$
($f$ is the integrand) Then taking $R \rightarrow \infty$ and $\epsilon \rightarrow 0$, should give me my result. But I happened to check out the integral in Wolfram Alpha and it gives the integral as $\frac{1}{2}\pi\sec\left(\frac{\pi}{8}\right)$, which is not what I get. Can I get some help? I'm sure I've gone wrong somewhere, and I'm pretty new at using these arguments, so help or insights will be nice. 

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\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{\infty}{\dd x \over
\root[{\large 4}]{x}\pars{1 + x^{2}}}}
\,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,
{1 \over 2}\int_{0}^{\infty}{x^{-5/8} \over 1 + x}\,\dd x
\\[5mm] \stackrel{x + 1\ \mapsto\ x}{=}\,\,\,&\
{1 \over 2}\int_{1}^{\infty}{\pars{x - 1}^{-5/8} \over x}\,\dd x
\,\,\,\stackrel{x\ \mapsto\ 1/x}{=}\,\,\,
{1 \over 2}\int_{1}^{0}{\pars{1/x - 1}^{-5/8} \over 1/x}\,
\pars{-\,{\dd x \over x^{2}}}
\\[5mm] = &\
{1 \over 2}\int_{0}^{1}x^{-3/8}\,\pars{1 - x}^{-5/8}\,\dd x =
{1 \over 2}\,{\Gamma\pars{5/8}\Gamma\pars{3/8} \over \Gamma\pars{1}} = {1 \over 2}\,{\pi \over \sin\pars{3\pi/8}}
\\[5mm] = &\
\bbx{{1 \over 2}\pi\sec\pars{\pi \over 8}} \approx 1.7002 \\ &
\end{align}
