Solve integrals using residue theorem? \int_0^{\pi}\frac{d\theta}{2+\cos\theta} \int_0^\infty\frac{x}{(1+x)^6}dx

For the second integral:
Note first that this integral is easily done by recognizing that ${\displaystyle x=(1+x)-1}$, so the integral is really
${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{dx}{{(1+x)}^5}-{\int }_0^{\mathrm{\infty }}\frac{dx}{{(1+x)}^6}=\frac{1}{4}-\frac{1}{5}=\frac{1}{20}}$
One may also use the residue theorem. However, one must choose an appropriate contour and integrand. In this case, a useful contour integral to consider is
${\displaystyle {\oint }_{C}dz\frac{z\log z}{{(1+z)}^6}}$
where C is a keyhole contour of outer radius R about the positive real axis. The contour integral is then equal to
${\displaystyle {\int }_{ϵ}^{R}dx\frac{x\log x}{{(1+x)}^6}+iR{\int }_0^{2\pi }d\theta e^{i\theta }\frac{R{e}^{i\theta }\log \left(R{e}^{i\theta }\right\}\left\{{(1+R{e}^{i\theta })}^6\right\}}{}}$
${\displaystyle +{\int }_R^{ϵ}dx\frac{x(\log x+i2\pi )}{{(1+x)}^6}+iϵ{\int }_{2\pi }^{0}d\varphi }$
${\displaystyle e^{i\varphi }\frac{ϵe^{i\varphi }\log \left(ϵe^{i\varphi }\right)}{{(1+ϵe^{i\varphi })}^6}}$
As ${\displaystyle R\to \mathrm{\infty }}$, the second integral vanishes as ${\displaystyle \frac{\log R}{R^4}}$. As ${\displaystyle ϵ\to 0}$, the fourth integral vanishes as ${\displaystyle {ϵ}^2\log ϵ}$. Thus, the contour integral is, in this limit
${\displaystyle -i2\pi {\int }_0^{\mathrm{\infty }}dx\frac{x}{{(1+x)}^6}}$
By the residue theorem, the contour integral is also equal to ${\displaystyle i2\pi }$ times the residue at the pole ${\displaystyle x=e^{i\pi }}$. (Note how important it is to get the argument correct.) The residue at this pole is
${\displaystyle \frac{1}{5!}{\left[\frac{d^5}{{dz}^5}\left(z\log z\right)\right]}_{z=e^{i\pi }}=-\frac{3!}{5!}}$
Putting this altogether, we get that
${\displaystyle {\int }_0^{\mathrm{\infty }}dx\frac{x}{{(1+x)}^6}=\frac{1}{20}}$
which agrees with the above.

Hint for the first one:
Consider the function
$f(z)=\frac{2}{z^2+4z+1}$
and find its poles. Then use the known formula for residues:
Under the assumption that f has a pole of order m at $z_0$
$Res(z_0,f)=\frac{1}{(m-1)!}\underset{z\to z_0}{lim}\frac{d^{m-1}}{d{z}^{m-1}}((z-z_0{)}^{m}f(z)).$
And finally, apply the Residue Theorem.