Evaluating the Definite Integral $\int_{0}^{1}\frac{2 \sin \pi x \cos \pi x}{1+x^2}dx$ 
How can I find this integral
  $$I=\int_{0}^{1}\frac{2 \sin \pi x \cos \pi x}{1+x^2}dx$$

Any trick that could compute the definite integral is acceptable. However, it will be more challenging to find a primitive.
Any hint or help is appreciated.

My Work
I just wrote the integral as
$$I=\int_{0}^{1}\frac{\sin 2 \pi x}{1+x^2}dx$$
Then, I decided to introduce
$$J(\alpha)=\int_{0}^{1}\frac{\sin \alpha x}{1+x^2}dx$$
and solve a more general problem. Hence, we would have $I=J(2\pi)$ as a special result. But I don't know how to go further. I tried integration by parts and substitutions but to no avail!

Let us consider $$I=\int \frac{\sin(2 \pi x)}{1+x^2} \,dx$$ Now, since $$\frac{1}{1+x^2}=\frac i 2\left( \frac{1}{x+i}-\frac{1}{x-i} \right)$$ hence $$I=\frac i 2\left(\int \frac{\sin(2 \pi x)}{x+i} \,dx-\int \frac{\sin(2 \pi x)}{x-i} \,dx\right)$$ Now, considering the first integral, writing $2\pi x=2\pi(x+i)-2i\pi$ and expanding $\sin(a-b)$ ,$$\sin(2 \pi x)=\cosh (2\pi ) \sin (2 \pi  (x+i))- i \sinh (2\pi)
    \cos (2 \pi  (x+i))$$ Now, change variable $2\pi(x+i)=t$ to make $$\int \frac{\sin(2 \pi x)}{x+i} \,dx=\cosh(2\pi)\int \frac{\sin(t)}{t} \,dt-i \sinh(2\pi)\int \frac{\cos(t)}{t} \,dt$$ and using $$\int \frac{\sin(t)}{t} \,dt=\text{Si}(t)\quad , \quad  \int \frac{\cos(t)}{t} \,dt=\text{Ci}(t)$$ Doing almost the same for the second integral, recombining and back to $x$ $$2I=\sinh (2 \pi ) (\text{Ci}(2 \pi  (i-x))+\text{Ci}(2 \pi  (i+x)))+i \cosh (2 \pi )
   (\text{Si}(2 \pi  (i+x))+\text{Si}(2 \pi (i- x))$$ from which you get  E.tsukerman's result.
Edit
Using the same approach for $$J=\int\frac{\sin (\alpha x)}{1+x^2}\,dx$$
$$2J=\sinh (\alpha ) (\text{Ci}(\alpha(i-x)  )+\text{Ci}(\alpha(i+x)  ))+i \cosh (\alpha )
   (\text{Si}(\alpha(i+x) )+\text{Si}( \alpha(i-x ))$$
Edit
Concerning the numerical value, it is awful (just as already said by other participants).
For $x=1$, the value is $\approx -0.077380457987+420.572696077718 i$
For $x=0$, the value is $\approx -0.170076586683+420.572696077718 i$
Edit
For the antiderivative, doing the same, it would have been faster to compute $$K=\int \frac{e^{i ax}}{1+x^2} \,dx=\frac i 2 \left( e^a \,\text{Ei}(a(ix-1))-e^{-a}\, \text{Ei}(a(i x+1))\right)$$
