<s>[INST] Computing $\int_{0}^{\infty} \frac{x^2 + a}{x^6 + a^3}dx$ I wish to compute

$$\int_{0}^{\infty} \frac{x^2 + a}{x^6 + a^3}dx, \quad a>0$$

but have no contour to work with. Does anyone have ideas on how to compute this integral?
 [/INST] This integral can be evaluated with calculus only:
\begin{align*}
\int_{0}^{\infty} \frac{x^2 + a}{x^6 + a^3} \, dx
&= \frac{1}{a^{3/2}} \int_{0}^{\infty} \frac{x^2 + 1}{x^6 + 1} \, dx \tag{1} \\
&= \frac{1}{a^{3/2}} \int_{0}^{\infty} \frac{1}{(x - \frac{1}{x})^2 + 1} \, \frac{dx}{x^2} \\
&= \frac{1}{2a^{3/2}} \int_{0}^{\infty} \frac{1+\frac{1}{x^2}}{(x - \frac{1}{x})^2 + 1} \, dx \tag{2} \\
&= \frac{1}{2a^{3/2}} \int_{-\infty}^{\infty} \frac{1}{u^2 + 1} \, du \quad (u = x-1/x) \\
&= \frac{\pi}{2a^{3/2}}.
\end{align*}
Explanation.


*

*For (1), we utilized the substitution $x \mapsto \sqrt{a} x$.

*For (2). the substitution $x \mapsto 1/x$ gives
$$ I := \int_{0}^{\infty} \frac{1}{(x - \frac{1}{x})^2 + 1} \, \frac{dx}{x^2} = \int_{0}^{\infty} \frac{1}{(x - \frac{1}{x})^2 + 1} \, dx. $$
Thus it follows that
$$ I = \frac{1}{2}(I + I) = \frac{1}{2} \int_{0}^{\infty} \frac{1 + \frac{1}{x^2}}{(x - \frac{1}{x})^2 + 1} \, dx. $$

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