Integrating $\int {1+\sin x\over x^2+2}dx$ I've been looking for a suitable approach to solve the following integral:

$$\int {1+\sin x\over x^2+2}dx$$

So far, I've tried by substitution and by parts to no avail. I've also attempted the following manipulations:
$$\int {\sin^2x+\cos^2x+\sin x\over x^2+2}dx = \int {\sin x (\sin x+1)+\cos^2 x\over x^2+2}dx $$
(Which appears to be just messing around with it; it got me not closer to any solution!)
And:
$$\begin{align}
\int {1+\sin x\over x^2+2}dx &= \int {(1+\sin x)(1-\sin x)\over (x^2+2)(1-\sin x)}dx \\
&= \int {1-\sin^2 x\over (x^2+2)(1-\sin x)}dx \\
&= \int {\cos^2(x)\over (x^2+2)(1-\sin x)}dx
\end{align}$$
However, I don't feel I'm getting any closer to a form I can solve. From what we are presently learning in our classes, we should be able to solve it using integration by parts, by substitution or by partial fractions. 
Could someone point me in the right direction so I can work it out?
Thans in advance.
Artem.
 A: $$\int {1+\sin(x)\over x^2+2}dx=\int {1\over x^2+2}dx+\int {\sin x \over x^2+2}dx$$
Here you can easily solve the first part Lets solve the 2nd part
$$I={\displaystyle\int}\dfrac{\sin\left(x\right)}{x^2+2}\,\mathrm{d}x={\displaystyle\int}\dfrac{\sin\left(x\right)}{\left(x-\sqrt{2}\mathrm{i}\right)\left(x+\sqrt{2}\mathrm{i}\right)}\,\mathrm{d}x$$
$$I={\displaystyle\int}\left(\dfrac{\mathrm{i}\sin\left(x\right)}{2^\frac{3}{2}\left(x+\sqrt{-2}\right)}-\dfrac{\mathrm{i}\sin\left(x\right)}{2^\frac{3}{2}\left(x-\sqrt{-2}\right)}\right)\mathrm{d}x$$
$$I=\class{steps-node}{\cssId{steps-node-1}{\dfrac{\mathrm{i}}{2^\frac{3}{2}}}}{\displaystyle\int}\dfrac{\sin\left(x\right)}{x+\sqrt{2}\mathrm{i}}\,\mathrm{d}x-\class{steps-node}{\cssId{steps-node-2}{\dfrac{\mathrm{i}}{2^\frac{3}{2}}}}{\displaystyle\int}\dfrac{\sin\left(x\right)}{x-\sqrt{2}\mathrm{i}}\,\mathrm{d}x$$
Now solving $I_1={\displaystyle\int}\dfrac{\sin\left(x\right)}{x+\sqrt{2}\mathrm{i}}\,\mathrm{d}x$
Substitute $u=x+\sqrt{2}\mathrm{i}$
$$I_1={\displaystyle\int}\dfrac{\sin\left(u-\sqrt{2}\mathrm{i}\right)}{u}\,\mathrm{d}u$$
$$I_1=\class{steps-node}{\cssId{steps-node-3}{\cosh\left(\sqrt{2}\right)}}{\displaystyle\int}\dfrac{\sin\left(u\right)}{u}\,\mathrm{d}u-\class{steps-node}{\cssId{steps-node-4}{\mathrm{i}\sinh\left(\sqrt{2}\right)}}{\cdot}{\displaystyle\int}\dfrac{\cos\left(u\right)}{u}\,\mathrm{d}u$$
$${\displaystyle\int}\dfrac{\sin\left(u\right)}{u}\,\mathrm{d}u=\operatorname{Si}\left(u\right)$$
$${\displaystyle\int}\dfrac{\cos\left(u\right)}{u}\,\mathrm{d}u=\operatorname{Ci}\left(u\right)$$
$$I_1=\cosh\left(\sqrt{2}\right)\operatorname{Si}\left(u\right)-\mathrm{i}\sinh\left(\sqrt{2}\right)\operatorname{Ci}\left(u\right)$$
$$I_1=\cosh\left(\sqrt{2}\right)\operatorname{Si}\left(x+\sqrt{2}\mathrm{i}\right)-\mathrm{i}\sinh\left(\sqrt{2}\right)\operatorname{Ci}\left(x+\sqrt{2}\mathrm{i}\right)$$
Now solving $I_2={\displaystyle\int}\dfrac{\sin\left(x\right)}{x-\sqrt{2}\mathrm{i}}\,\mathrm{d}x$
apply the same method you will get
$$I_2=\cosh\left(\sqrt{2}\right)\operatorname{Si}\left(x-\sqrt{2}\mathrm{i}\right)+\mathrm{i}\sinh\left(\sqrt{2}\right)\operatorname{Ci}\left(x-\sqrt{2}\mathrm{i}\right)$$
Putting back values you get 
$$I=\dfrac{\mathrm{i}\cosh\left(\sqrt{2}\right)\left(\operatorname{Si}\left(x+\sqrt{2}\mathrm{i}\right)-\operatorname{Si}\left(x-\sqrt{2}\mathrm{i}\right)\right)+\sinh\left(\sqrt{2}\right)\left(\operatorname{Ci}\left(x+\sqrt{2}\mathrm{i}\right)+\operatorname{Ci}\left(x-\sqrt{2}\mathrm{i}\right)\right)}{2^\frac{3}{2}}+C$$
A: $$I=\int\frac{1+\sin(x)}{x^2+2}dx=\int\frac{1}{x^2+2}dx+\int\frac{\sin(x)}{x^2+2}dx$$
$$I_1=\int\frac{1}{x^2+2}dx=\int\frac{1}{2\left[(\frac{x}{\sqrt{2}})^2+1\right]}dx$$
$a=\frac{x}{\sqrt{2}}$ so $dx=\sqrt{2}da$
$$I_1=\frac{\sqrt{2}}{2}\int\frac{1}{a^2+1}da=\frac{\sqrt{2}}{2}\arctan\left(\frac{x}{\sqrt{2}}\right)+C$$
now we can move onto the harder second integral:
$$I_2=\int\frac{\sin(x)}{x^2+2}dx=\Im\int\frac{e^{ix}}{x^2+2}dx$$
if we let:
$$I_2(b)=\Im\int\frac{e^{iax}}{x^2+2}dx$$
$$I_2'(b)=\Im\int\frac{ix.e^{iax}}{x^2+2}dx$$
$u=x^2+2$ so $dx=\frac{du}{2x}$
$$I_2'(b)=\Im\frac{i}{2}\int\frac{e^{ia\sqrt{u-2}}}{u}du=\Im\frac{i}{2}\int\frac{1}{u}\sum_{n=0}^{\infty}\frac{(ia\sqrt{u-2})^n}{n!}du$$
and this $\sqrt{u-2}$ could also be expressed as a series to give a double summation which may be solvable. 
