Paths integrals in the complex plane

For this function:

f[z_] := (1 - E^z + z)/(z^3 (z - 1)^2) 

there are no branch cuts in the complex plane therefore we simply use Cauchy integral theorem and the related formula of the complex residue, i.e. we sum up residues of the function $f$ in the circle $\mid z \mid =2$. Let's denote $$int = \oint_{\mid z \mid =2}\frac{1-e^z+z}{z^3 (z-1)^2}dz$$ Now we have:

int = 2 Pi I Total[ Residue[f[z], {z, #}] & /@ {0, 1}] // Simplify 

 I (-11 + 4 E) Pi 

Alternatively we can parametrize z over the given circle z -> 2 E^(I t):

(1 - E^z + z)/(z^3 (z - 1)^2) /. z -> 2 E^(I t) 

(E^(-3 I t) (1 - E^(2 E^(I t)) + 2 E^(I t)))/(8 (-1 + 2 E^(I t))^2) 

and d z -> 2 I E^(I t) d t, now we have:

Integrate[(E^(-3 I t)(1 - E^(2 E^(I t)) + 2 E^(I t)))/( 8(-1 + 2 E^(I t))^2) 2 I E^(I t), {t, 0, 2 Pi}] 

I (-11 + 4 E) Pi 

% // TraditionalForm 

If a numerical answer is good enough you can just enter the path. As @Artes said it doesn't have to be the circle exactly.

NIntegrate[f[z], {z, 2 - 2 I, 2 + 2 I, -2 + 2 I, -2 - 2 I, 2 - 2 I}] (* 0. - 0.398582 I *) 

Check :

I (-11 + 4 E) Pi // N (* 0. - 0.398582 I *) 

Another suggestion from @Artes (thanks !) : one can use symbolic integration as well and

Integrate[f[z], {z, 2 - 2 I, 2 + 2 I, -2 + 2 I, -2 - 2 I, 2 - 2 I}] // FullSimplify 

will reproduce his result.