Your integral is
$$ \int_0^{\infty} \frac{x^{7/5}}{1+x^2} \frac{dx}{x}. $$
Substituting $u=x^2$, $du/u=2dx/x$, the integral becomes
$$ \frac{1}{2}\int_0^{\infty} \frac{u^{7/10}}{1+u} \, du $$
Now,
$$ \frac{1}{1+u}= \int_0^{\infty} e^{-(1+u)\alpha} \, d\alpha, $$
and interchanging the order of integration gives
$$ \frac{1}{2}\int_0^{\infty} e^{-\alpha} \left( \int_0^{\infty} u^{7/10-1} e^{-\alpha u} \, du \right) \, d\alpha $$
Changing variables in the inner integral shows it has value $\alpha^{-7/10}\Gamma(7/10)$. We then do
$$ \frac{\Gamma(7/10)}{2} \int_0^{\infty} \alpha^{3/10-1} e^{-\alpha} \, d\alpha = \frac{\Gamma(7/10)}{2}\Gamma(3/10) = \frac{\pi}{2\sin{(3\pi/10)}}, $$
which you do by whatever arcane trigonometry you have to hand.
On the other hand, there is a more direct way of getting the relation
$$ \int_0^{\infty} \frac{x^{s-1}}{1+x} \, dx = \frac{\pi}{\sin{\pi s}}, \quad 0<\Re(s)<1. $$
Split the integral in two at $x=1$:
$$ \int_0^{1} \frac{x^{s-1}}{1+x} \, dx + \int_1^{\infty} \frac{x^{s-1}}{1+x} \, dx. $$
Now set $u=1/x$ in the second integral, and we find
$$ \int_0^{1} \frac{x^{s-1}}{1+x} \, dx + \int_0^{1} \frac{u^{1-s-1}}{1+u} \, du = \int_0^1 \frac{x^{s-1}+x^{-s}}{1+x} \, dx. $$
And now we use a trick I saw in some of G.H. Hardy's work: expand the denominator in a power series and change the order of integration (we can check this is legal easily enough):
$$ \int_0^1 \frac{x^{s-1}+x^{-s}}{1+x} \, dx = \sum_{n=0}^{\infty} (-1)^n \int_0^{\infty} (x^{n+s-1}+x^{n-s}) \, dx $$
Doing the integrals gives
$$ \int_0^1 \frac{x^{s-1}+x^{-s}}{1+x} \, dx = \sum_{n=0}^{\infty} (-1)^n \left( \frac{1}{n+s} + \frac{1}{n-s+1} \right) $$
Shifting the second terms by one (which you can check is okay by trucating the sums, we end up with
$$ \int_0^{\infty} \frac{x^s}{1+x} \, dx = \frac{1}{s} + \sum_{n=1}^{\infty} (-1)^n \left(\frac{1}{s+n}+\frac{1}{s-n} \right), $$
which is a well-known expression for $\pi\csc{\pi s}$. (Which is actually what Hardy says. You can get it using Fourier series, and so avoid both complex analysis and the Gamma function entirely.)