Two nice integrals: $\int_{-\infty}^{+\infty} \frac{\cos(x^2)}{1 + x^2} dx$ and $\int_{-\infty}^{+\infty} \frac{\sin(x^2)}{1 + x^2} dx$ Wolfram Alpha solve those two integrals, and apparently it depends from the residue of the pole at $x = i$ and at infinity, as far I could solve.
The results can be previewed here:
https://www.wolframalpha.com/input/?i=integrate(sin(x%5E2)%2F(1%2Bx%5E2),x,-inf,%2Binf)
https://www.wolframalpha.com/input/?i=integrate(cos(x%5E2)%2F(1%2Bx%5E2),x,-inf,%2Binf)
Part of the answer makes sense, since it can be directly evaluated by the Residue Theorem.
However the terms that depends from the Fresnel Functions are a little strange.
How to solve this integral that combine a Fresnel Integral and a rational function that should be solvable using the Residues Theorem ?
 A: Using the tip giving by J.G. the problem can be view as the following complex integral:
$ f(s) = \int_{-\infty}^\infty \frac{\exp{(i s (1 + x^2))}}{1 + x^2} dx$
Taking the derivative of $f(s)$, it will give a Complex Fresnel Integral:
$ f'(s) = \int_{-\infty}^\infty i \exp{(i s (1 + x^2))} dx$
Luckily, the solution for this integral was already solved on standard textbooks, so I will skip this step:
$ f'(s) = \sqrt{\frac{\pi}{2 s}} (i + 1) i \exp{(i s)}$
Finally, the original problem are almost equal to $f(1)$, which implies to evaluate the integral:
$ f(1) - f(0) = \int_0^1 \sqrt{\frac{\pi}{2 s}} (i + 1) i \exp{(i s)} ds$
Where $f(0) = \int_{-\infty}^\infty \frac{1}{1 + x^2} dx = \pi $ is a trivial integral
The integral itself, after a simple substitution that I don't reproduce here, should get the Euler's error function. When I use Maxima and simplify the expression, luckily, it gets a simple formula:
$ f(1) = \pi - \pi \times erf{(\sqrt{-i})}$
By the other hand, the initial function are also represented by:
$ f(1) = \int_{-\infty}^\infty \frac{\exp{(i(1 + x^2))}}{1 + x^2} dx = exp{(i)} \int_{-\infty}^\infty \frac{\exp{(i x^2)}}{1 + x^2} dx$
Joining together, it will get:
$\int_{-\infty}^\infty \frac{\exp{(i x^2)}}{1 + x^2} dx = \pi \exp{(-i)} - \pi \exp{(-i)} \times erf{(\sqrt{-i})}$
Finally, the final results will be:
$\int_{-\infty}^\infty \frac{\cos{(x^2)}}{1 + x^2} dx = \Re (\pi \exp{(-i)} - \pi \exp{(-i)} \times erf{(\sqrt{-i})} ) \approx 1.305608$
$\int_{-\infty}^\infty \frac{\sin{(x^2)}}{1 + x^2} dx = \Im (\pi \exp{(-i)} - \pi \exp{(-i)} \times erf{(\sqrt{-i})} ) \approx 0.723571$
Using the reference from Wikipedia about Fresnel Functions are Error Function, luckily, we obtain:
$ C(1) + i S(1) = \sqrt{\frac{\pi}{2}} \frac{(1 + i)}{2} erf(\sqrt{- i}) $
Where:
$ C(1) = \int_0^1 \cos(x^2) dx $
$ S(1) = \int_0^1 \sin(x^2) dx $
And the real and imaginary parts of the Error's function can be separated, resulting:
$\int_{-\infty}^\infty \frac{\cos{(x^2)}}{1 + x^2} dx = \pi \cos{(1)} - \frac{\sqrt{8 \pi} [S(1)(\sin(1) + \cos(1)) + C(1)(\cos(1)-\sin(1))]}{2}$
$\int_{-\infty}^\infty \frac{\sin{(x^2)}}{1 + x^2} dx = -\pi \sin{(1)} - \frac{\sqrt{8 \pi} [S(1)(\cos(1) - \sin(1)) - C(1)(\cos(1)+\sin(1))]}{2}  $
