Can I integrate this without using Cauchy's integral formula How can I integrate this?
$$\frac{1}{2 \pi i} \oint_c \frac{e^z}{z(1-z)^3}dz $$
C is any simple path which encloses 0 and 1. 
And can it be done without using Cauchy's integral formula?
 A: You can write $$ \dfrac{e^z}{z(1-z)^2} = \dfrac{1}{z} + \dfrac{e^z-1}{z} + \dfrac{d}{dz} \left(\dfrac{e^z}{1-z}\right) $$
The last term is explicitly the derivative of a meromorphic function, so its integral around any contour that avoids $z=1$ is $0$.  The singularity of the second term is removable, so its integral around any contour is $0$.  That leaves the first term, which you can explicitly integrate (or use the fact that it has antiderivative $\log z$ which jumps by $2\pi i$ when you cross the branch cut).
A: Take little circles around $\;z=0,\,1\;$ , say $\;C_0,\;C_1\;$  of radius $\;0.1\;$, so the integral equals the sum of the integrals around these two circles . Now Laurent series around these points, using the well known ones for the exponential function and the geometric series:
$$\begin{align*}&z=0:\;\;\frac1z\frac{e^z}{(1-z)^3}=\frac1z\left(1+z+\frac{z^2}2+\ldots\right)\left(1+z+z^2+\ldots\right)^3=\frac1z\left(1+\ldots\right)=\\{}\\&=\frac1z+\ldots\implies \text{Res}_{z=0}(f)=1\\{}\\&z=1:\;\;-\frac1{(z-1)^3}\cdot\frac{e\cdot e^{z-1}}{1+(z-1)}=\\{}\\&=-\frac e{(z-1)^3}\cdot\left(1+(z-1)+\frac{(z-1)^2}2+\ldots\right)\left(1-(z-1)+(z-1)^2-\ldots\right)=\\{}\\&=-\frac e{(z-1)^3}\left(1+\frac{(z-1)^2}2+\ldots\right)=...-\frac e{2(z-1)}+\ldots\implies\;\text{Res}_{z=1}(f)=-\frac e2\end{align*}$$
Thus we get that
$$\frac1{2\pi i}\int_C\frac{e^z}{z(1-z)^3}dz=\frac1{2\pi i}\left(\int_{C_1}\frac{e^z}{z(1-z)^3}dz+\int_{C_2}\frac{e^z}{z(1-z)^3}dz\right)=\frac1{2\pi i}\left(1-\frac e2\right)$$
