Evaluating $\int \frac{\sin\left(x\right)}{1+x^2}dx$

$$\int \frac{\sin\left(x\right)}{1+x^2}dx$$

I have tried to integrate by parts but it doesn't work. How do I evaluate it? Any advice, hint or well-thought solution will be appreciated.

@Lucian gave very good explanations and reasons for which this integral cannot be computed at least in terms of elementary functions.

Using, as I commented ealier, $$\frac 1 {1+x^2}=\frac i2 \Big(\frac 1 {x+i}-\frac 1 {x-i}\Big)$$ the only result I was able to get (after a few simplifications) is $$I=\int \frac{\sin\left(x\right)}{1+x^2}dx=\frac{\left(e^2-1\right)\Big( \text{Ci}(x+i)+ \text{Ci}(x-i)\Big)+i \left(e^2+1\right) \Big(\text{Si}(x+i)+\text{Si}(x-i)\Big)}{4 e}$$ where appear the sine and cosine integrals.

I did not find any way to simplify this expression.

Edit

To show how this works, let us consider $$J=\int \frac{\sin(x)}{x+i}dx$$Change variable $x=y-i$; so $$J=\int \frac{\sin(y+i)}{y}dy$$ Developing the sine and using the fact that $\sin(i)=i \sinh (1)$, $\cos(i)= \cosh (1)$, we then have $$J=\int \Big(\frac{\cosh (1) \sin (y)}{y}-\frac{i \sinh (1) \cos (y)}{y}\Big)\,dy$$ that is to say $$J=\cosh (1) \,\text{Si}(y)-i \sinh (1) \,\text{Ci}(y)$$

Similarly, $$K=\int \frac{\sin(x)}{x-i}dx$$ Change variable $x=z+i$; so $$K=\int \frac{\sin(z-i)}{z}dz$$ $$K=\int \Big( \frac{\cosh (1) \sin (z)}{z}+\frac{i \sinh (1) \cos (z)}{z}\Big)\,dz$$ that is to say $$K=\cosh (1)\, \text{Si}(z)+i \sinh (1)\, \text{Ci}(z)$$ and finally $$I=\int \frac{\sin(x)}{1+x^2}dx=\frac i2(J-K)$$

How do I evaluate it?

You don't, because it can't be done. At least not in terms of elementary functions. The same goes

for $\displaystyle\int\frac{\cos(x)}{1+x^2}~dx$ as well, the difference being that the latter has a beautiful definite counterpart,

namely $\displaystyle\int_{-\infty}^\infty\frac{\cos(x)}{1+x^2}~dx~=~\frac\pi e$ . Of course, there's also $\displaystyle\int_{-\infty}^\infty\frac{\sin^2(x)}{1+x^2}~dx~=~\frac\pi2~\Big(1-e^{-2}\Big)$. The

reason for this is the parity of the integrand.