Does this residue trick work? When I have to evaluate an integral by means of residues I sometimes find this shortcut useful. Let's say I need to integrate
$$I = \int_{-\infty}^{+\infty} \frac{x^2}{x^4+6x^2+13}dx$$
I would normally go with the starndard procedure of taking the limit of the integral on the semincircle $[-R,R] \cup \{Re^{it} : t \in [0,\pi)\}$ in the upper plane and evaluate it with residues.
Though if I make the substituition $y = x^2$ I get
$$I = \frac{1}{2}\int_{-\infty}^{+\infty} \frac{\sqrt y}{y^2+6y+13}dy$$
which is far easier to compute by the same procedure with residues. The integrand though is not a rational function anymore, so the question is: am I allowed to compute the second integral with the semicircle path as as I would do for the first one?
I'd say it's legitimate, the integrand on the arc $Re^{it}$ tends to zero as R tends to intinity and the poles are just determined by the denominator, which is still a polinomial.
By the way using this method I get $I = \pi \sqrt{-3+2i}/8$
 A: The new function is not analytic. It is not even differentiable at $0$. So, you cannot use the residue theorem.
A: Your final answer of $\pi\sqrt{-3+2i}/8$ answers your question.
The integral is clearly a positive real number, but
$\pi\sqrt{-3+2i}/8$ isn't real.
A: Your substitution is wrong. 
$$ \int_{0}^{+\infty} \frac{x^2}{x^4+6x^2+13}dx
= \frac{1}{2}\int_{0}^{+\infty} \frac{\sqrt y}{y^2+6y+13}dy $$
$$ \int_{-\infty}^{0} \frac{x^2}{x^4+6x^2+13}dx
= \frac{1}{2}\int_{+\infty}^{0} \frac{-\sqrt y}{y^2+6y+13}dy
= \frac{1}{2}\int_{0}^{+\infty} \frac{\sqrt y}{y^2+6y+13}dy
$$
A: As others have already mentioned, the substitution in the OP is invalid.  Instead, we exploit the evenness of the integrand and write
$$\begin{align}
I&=2\int_0^\infty \frac{x^2}{x^4+6x^2+13}\,dx\\\\
&=\int_0^\infty \frac{\sqrt{x}}{x^2+6x+13}\,dx\tag1
\end{align}$$
We can evaluate the integral on the right-hand side of $(1)$ using complex analysis.  
Let $f(z)=\frac{\sqrt z}{z^2+6z+13}$ for $z\in \mathbb{C}\setminus [0,\infty)$ with $0<\arg(z)\le 2\pi$ (i.e., We have cut the complex plane along the positive real axis).  
Let $C$ be the classical "keyhole" contour where the keyhole coincides with the branch cut.
Then, we have
$$\begin{align}
\int_0^\infty \frac{\sqrt{x}}{x^2+6x+13}\,dx&=i\pi \,\text{Res}\left(\frac{\sqrt z}{z^2+6z+13}, z=-3\pm i4\right)\\\\
&=i\pi\,\left(\frac{\sqrt {13}e^{\frac i2\text{atan2}(2,-3)}}{4i}-\frac{\sqrt {13}e^{\frac i2\left(2\pi+\text{atan2}(-2,-3)\right)}}{4i}\right)\\\\
&=\frac{\sqrt {13}\pi}{2}\cos\left(\frac12 \text{atan2}(2,-3)\right)\\\\
&=\frac{\pi}{2}\sqrt{\frac{\sqrt{13}-3}{2}}
\end{align}$$
A: You can split the integral in partial fractions
$$\frac{\frac{1}{2}+\frac{3 i}{4}}{x^2+(3-2 i)}+\frac{\frac{1}{2}-\frac{3 i}{4}}{x^2+(3+2 i)}$$
Then compute the residues for each part
$$\frac{\frac{1}{2}+\frac{3 i}{4}}{2 \sqrt{-3+2 i}};\quad \frac{-\left(\frac{1}{2}-\frac{3 i}{4}\right)}{2 \sqrt{-3-2 i}}$$
and get
$$2 \pi  i \left(\frac{\frac{1}{4}+\frac{3 i}{8}}{\sqrt{-3+2 i}}-\frac{\frac{1}{4}-\frac{3 i}{8}}{\sqrt{-3-2 i}}\right)$$
which simplifies to
$$\frac{1}{2}\, \pi \sqrt{\frac{1}{2} \left(\sqrt{13}-3\right)} $$
