The poles are clearly 1 and 2. $C$ is the circle with equation $(x+1)^2+(y-1)^2=4$. Putting $y=0$, we get $x=\sqrt{3} -1, -\sqrt{3} - 1$. This means both poles lie outside the circle. So both residues are zero. The integral is $2\pi i$ times sum of residues, so it's also zero.
Is my work correct?
For a rational function with pole $0$,you simply obtain its Laurent's expansion dividing by increasing powers its numerator by its denominator.
For a pole $\alpha\ne 0$, you first use the substitution $u=z-\alpha$, to obtain a rational function in $u$, with pole $0$.
I'll show how it runs for the pole $2$ of $f(z)$. So we set $u=z-2$ and $$f(z)=\frac{2+u}{(1+u)u^2}=g(u).$$ Now to the division:
\begin{alignat}{4} \dfrac2{u^2} &-\dfrac 1u&{}+{}&1&{}-{}&u +\dotsm \\ u^2+u^3\Bigm( \phantom{-}2&+u \\ -2&-2u \\[-2ex] - &--\\[-2ex] &-u\\ &+u &{}+{}&u^2 \\[-1ex] &---&-&-\\[-1ex] &&{}-{}&u^2&{}-{}&u^3\\[-1ex] &&&---&-\\[-1ex] &&&&{}-{}&u^3\\[-1ex] &&&&&\ddots \end{alignat} Therefor the residue at pole $2$ is equal to $-1$.