Note: we're not yet allowed to say derivative of holomorphic are holomorphic or even that they are continuous. hell.
For $\int_{\gamma}z^2e^{z^3}dz$, where $\gamma:[0,1] \to \mathbb C$, $\gamma(t)=(t-1)^2+it^3$
For independence of path theorem below, choose
$$(D,a,b,f(z),F(z)) =(\mathbb C,\gamma(0)=1,\gamma(1)=i,z^2e^{z^3},\frac{e^{z^3}}{3})$$
- Let $a,b \in \mathbb C$. Let $f: D \to \mathbb C$ be continuous for open and connected $D \subseteq \mathbb C$. Let $\gamma$ be any piecewise smooth path starting at $a$ and ending at $b$. Suppose $f$ has a primitive/an antiderivative $F$ in $D$. Then $\int_\gamma f = F(b)-F(a)$.
We can do this because
$\gamma$ is a piecewise smooth since $\gamma$ is a smooth path since both its real and imaginary parts are smooth since both parts are polynomials.
$f(z)=z^2e^{z^3}$ is continuous in $D=\mathbb C$ because $f$ is entire because $f$ is the product of 2 entire functions where $e^{z^3}$ is entire because it is the composition of 2 entire functions $e^z$ and $z^3$.
Therefore, (see wolfram here and here)
$$\int_{\gamma}z^2e^{z^3}dz$$
$$ = \int_\gamma f = \int_\gamma f(z) dz = F(b)-F(a)$$
$$= \frac{e^{i^3}}{3} - \frac{e^{1^3}}{3}$$
$$=\frac{-e}{3} + \frac{e^{-i}}{3}$$
$$= \frac{[-e+\cos(1)]+i[-\sin(1)]}{3}$$