<s>[INST] 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.
 [/INST] 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}
</s>