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's try your last idea and start with :
$$J(\alpha):=\int_{0}^{1}\frac{\sin \alpha x}{1+x^2}dx$$
Then $\;\displaystyle J''(\alpha)=-\int_{0}^{1}\frac{x^2\;\sin \alpha x}{1+x^2}dx\;$ and we obtain this ODE :
$$J(\alpha)-J''(\alpha)=\frac {1-\cos(\alpha)}{\alpha}$$
The solution of the homogeneous ODE is simply $\;J(x)=ae^x+be^{-x}$.
Let's use variation of constant and start with $J(x)=a(x)e^x$ then :
\begin{align}
J(x)&=a(x)e^x\\
J'(x)&=(a'(x)+a(x))e^x\\
J''(x)&=(a''(x)+2a'(x)+a(x))e^x\\
J(x)-J''(x)&=-(a''(x)+2a'(x))e^x=\frac {1-\cos(x)}{x}\\
\end{align}
For $b(x):=-a'(x)\,$ we have 
\begin{align}
b'(x)+2b(x)&=\frac {1-\cos(x)}{x}e^{-x}\\
\end{align}
For $b(x):=c(x)e^{-2x}\,$ this becomes for $\operatorname{Ei}$ the exponential integral :
\begin{align}
c'(x)&=\frac {1-\cos(x)}{x}e^{x}\\
c(x)&=\int\frac {e^{x}}x\,dx-\int\frac {e^{x+ix}+e^{x-ix}}{2\,x}\,dx\\
c(x)&=C+\operatorname{Ei}(x)-\frac 12\operatorname{Ei}((1+i)x)-\frac 12\operatorname{Ei}((1-i)x)\\
\end{align}
Coming back to $a(x)$ :
\begin{align}
a'(x)&=-\left(C+\operatorname{Ei}(x)-\frac 12\operatorname{Ei}((1+i)x)-\frac 12\operatorname{Ei}((1-i)x)\right)e^{-2\,x}\\
\end{align}
Now $\ \displaystyle \int \operatorname{Ei}(u\,x)e^{-vx}\,dx=\frac 1v\left(\operatorname{Ei}((u-v)x)-e^{-vx}\operatorname{Ei}(u x)\right)\;$ from Wolfram functions so that for $\,v=2$ and $u=1,1+i,1-i$ :
$$a(x)=D-\frac 12\left(C_1\,e^{-2x}+\operatorname{Ei}(-x)-e^{-2x}\operatorname{Ei}(x)\\-\frac 12\left(\operatorname{Ei}((-1+i)x)-e^{-2x}\operatorname{Ei}((1+i)x)+\operatorname{Ei}((-1-i)x)-e^{-2x}\operatorname{Ei}((1-i)x)\right)\right)$$
Multiplying by $e^x$ we get the general formula for $J(x)$ :
$$J(x)=De^x+D_1\,e^{-x}-\frac {e^x\operatorname{Ei}(-x)-e^{-x}\operatorname{Ei}(x)}2\\+\frac {e^x\operatorname{Ei}((-1+i)x)-e^{-
x}\operatorname{Ei}((1+i)x)+e^x\operatorname{Ei}((-1-i)x)-e^{-x}\operatorname{Ei}((1-i)x)}4$$
(of course getting this with alpha is faster!)
Now we have to find $J(2\pi)$ knowing that $J(0)=0$ (I'll let you try too!).
