Definite integral (mixture of functions) I have problem with this integral and I generally don't know how to approach it:
$$\int_{-2}^2 (x^4+4x+\cos(x))\cdot \arctan\left(\frac{x}{2}\right)dx$$
I know that I probably have to make some substitution, but really don't know which.
Thanks for your help.

Using lab bhattacharjee's result. The integral can be solven by using integration by parts.
Let
\begin{align}
u=\arctan\frac{x}{2}\qquad\rightarrow\qquad du=\frac{2\,dx}{x^2+4}\qquad\text{and}\qquad dv=x\,dx\qquad\rightarrow\qquad v=\frac{1}{2}x^2
\end{align}
Then
\begin{align}
8\int_0^2x\arctan\frac {x}{2}\, dx&=\left.4x^2\arctan\frac{x}{2}\right|_0^2-8\int_0^2 \frac{x^2}{x^2+4}\,dx\\
&=4\pi-8\int_0^2 \frac{x^2+4-4}{x^2+4}\,dx\\
&=4\pi-8\int_0^2 \,dx+16\int_0^2 \frac{2}{x^2+4}\,dx\\
&=4\pi-16+16\, \left.\arctan\frac{x}{2}\right|_0^2\\
&=4\pi-16+4\pi\\
&=8\pi-16
\end{align}
