$\int_{0}^{\infty} \frac {\sqrt x}{x^2-5x+6} \,dx = \frac {\pi}{i \sqrt 3 +i \sqrt 2}$ Prove
$$\int_{0}^{\infty} \frac {\sqrt x}{x^2-5x+6} \,dx = \frac {\pi}{i \sqrt 3 +i \sqrt 2}$$
[Nevermind!]: It has been said, it can be solved in at least three ways.I'm looking forward to seeing two other ways.
 A: The given result is wrong, if the integral is interpreted as a principal value integral. It doesn't make sense for any natural interpretation of an integral that the value of a real integral should be a nonzero purely imaginary number anyway, so also for other interpretations, the given result is most likely wrong.
One way to compute the integral (as a principal value integral) is via complex analysis, using the residue theorem:
Take a keyhole contour $C_{\varepsilon,R}$ enveloping the positive real half-axis, with circular deviations around the poles of the integrand in $2$ and $3$ - similar to 
but with two circular deviations for the poles. (Image taken from this question about an integral of a function with a pole.)
Since the contour encloses no singularity of the function, the integral along that contour is $0$. The integrals along the circular parts of the contour tend to $0$ as the radius $\varepsilon$ of the small circular arc around $0$ shrinks to $0$ and the radius $R$ of the large circular arc tends to $+\infty$ by the standard estimate. Since the value of $\sqrt{z}$ as $z$ approaches the positive real half-axis from the lower half plane is the negative of the value as $z$ approaches the positive half-axis from the upper half-plane, the signs from the different orientation on the real axis are cancelled by the sign of the square root, and what remains is
$$\begin{align}
0 &= 2 \int_{M(\varepsilon)} \frac{\sqrt{x}}{x^2-5x+6}\,dx\\
&\qquad + \int_0^\pi \frac{\sqrt{2-\varepsilon e^{-it}}}{(-\varepsilon e^{-it})(-\varepsilon e^{-it}-1)}\,d(2-\varepsilon e^{-it}) + \int_0^\pi \frac{\sqrt{3-\varepsilon e^{-it}}}{(1-\varepsilon e^{-it})(-\varepsilon e^{-it})}\,d(3-\varepsilon)\\
&\qquad + \int_0^\pi \frac{\sqrt{2 + \varepsilon e^{-it}}}{\varepsilon e^{-it}(\varepsilon e^{-it}-1)}\, d(2+\varepsilon e^{-it}) + \int_0^\pi \frac{\sqrt{3+\varepsilon e^{-it}}}{(1+\varepsilon e^{-it})\varepsilon e^{-it}}\, d(3+\varepsilon e^{-it}),
\end{align}$$
where $M(\varepsilon) = [0,2-\varepsilon] \cup [2+\varepsilon,3-\varepsilon] \cup [3+\varepsilon, \infty)$.
Now we have to note that for the semicircles in the lower half-plane (the third and fourth of the semicircle integrals) the square root has negative real part, so
$$\begin{align}
\int_0^\pi \frac{\sqrt{2-\varepsilon e^{-it}}}{(-\varepsilon e^{-it})(-\varepsilon e^{-it}-1)}\,d(2-\varepsilon e^{-it})&\; + \int_0^\pi \frac{\sqrt{2 + \varepsilon e^{-it}}}{\varepsilon e^{-it}(\varepsilon e^{-it}-1)}\, d(2+\varepsilon e^{-it})\\
&= i\int_0^\pi \frac{\sqrt{2-\varepsilon e^{-it}}}{1+\varepsilon e^{-it}} + \frac{\sqrt{2+\varepsilon e^{-it}}}{1-\varepsilon e^{-it}}\,dt
\end{align}$$
tends to $0$ as $\varepsilon \to 0$, and similarly for the semicircles around $3$, whence we obtain
$$\operatorname{v.p.} \int_0^\infty \frac{\sqrt{x}}{x^2-5x+6}\,dx = 0.$$
Another way to compute the same result is to determine a local primitive of the integrand on each of the intervals $(0,2)$, $(2,3)$, and $(3,\infty)$. Since
$$\frac{\sqrt{x}}{x^2-5x+6} = \frac{\sqrt{x}}{(x-3)(x-2)} = \frac{\sqrt{x}}{x-3} - \frac{\sqrt{x}}{x-2},$$
a little fiddling leads to the primitive
$$\sqrt{3}\log \left\lvert \frac{\sqrt{x}-\sqrt{3}}{\sqrt{x}+\sqrt{3}}\right\rvert - \sqrt{2} \log \left\lvert \frac{\sqrt{x}-\sqrt{2}}{\sqrt{x}+\sqrt{2}}\right\rvert$$
on each of the above intervals. Then one can compute the integrals over the intervals $(\varepsilon,2-\varepsilon)$, $(2+\varepsilon,3-\varepsilon)$ and $(3+\varepsilon,\infty)$ explicitly, sum them, and take the limit as $\varepsilon \to 0$ to once again obtain
$$\operatorname{v.p.} \int_0^\infty \frac{\sqrt{x}}{x^2-5x+6}\,dx = 0.$$
A: One possible approach is using the residue theorem.  First make the change of variable $x = u^2$ then we have
$$\int _{0}^{\infty }\!{\frac {\sqrt {x}}{{x}^{2}-5\,x+6}}{dx}=2\,\int _
{0}^{\infty }\!{\frac {{u}^{2}}{{u}^{4}-5\,{u}^{2}+6}}{du}=\int _{-
\infty }^{\infty }\!{\frac {{u}^{2}}{{u}^{4}-5\,{u}^{2}+6}}{du}
$$
The last integral is converted into a contour integral and it is evaluated using the residue theorem:
$$\int _{-\infty }^{\infty }\!{\frac {{u}^{2}}{{u}^{4}-5\,{u}^{2}+6}}{du
}=2\,i\pi \, \left( {\it Res} \left( {\frac {{u}^{2}}{{u}^{4}-5\,{u}^{
2}+6}},u=-\sqrt {3} \right) +{\it Res} \left( {\frac {{u}^{2}}{{u}^{4}
-5\,{u}^{2}+6}},u=-\sqrt {2} \right)+{\it Res} \left( {\frac {{u}^{2}}{{u}^{4}
-5\,{u}^{2}+6}},u=\sqrt {2} \right)+{\it Res} \left( {\frac {{u}^{2}}{{u}^{4}
-5\,{u}^{2}+6}},u=\sqrt {3} \right)  \right)$$
Computing the residues we obtain
$$\int _{0}^{\infty }\!{\frac {\sqrt {x}}{{x}^{2}-5\,x+6}}{dx}=0
$$
This result was checked using Mathematica with the command
Integrate[Sqrt[x]/(x^2 - 5*x + 6), {x, 0, ∞}, PrincipalValue -> True]

Thanks to @DanielFischer by the corrections.
