Let $G \subset \mathbb C$ be an open set, $f: G \to \mathbb C$ a continuous function and $\gamma: [\alpha, \beta] \to G$ a contour. Define the contour integral of $f$ along $\gamma$ to be \begin{align*} \int_\gamma f = \sum_{k = 1}^n \int_{t_{k-1}}^{t_k} f(\gamma(t)) \gamma'(t)dt, \end{align*} where $\alpha = t_0 < t_1 < \ldots < t_n = \beta$ is the partition of $[\alpha, \beta]$, associated to $\gamma = \sum_{i = 1}^n \gamma_i$.
I tried to prove the (simple) fact that, if we have a smooth contour then
\begin{align*} \int_{\sum_{i = 1}^n \gamma_i} f = \sum_{i = 1}^n \int_{\gamma_i} f. \end{align*}
I suppose it is straight forward, just with the definition, and I reached the following point: \begin{align*} \int_{\sum_{i = 1}^n \gamma_i} f &= \int_\alpha^\beta f\left(\sum_{i = 1}^n \gamma_i(t)\right) \left(\sum_{i = 1}^n \gamma_i(t)\right)' dt = \sum_{i = 1}^n \int_\alpha^\beta f\left(\sum_{i = 1}^n \gamma_i(t)\right) \gamma_i'(t) dt. \end{align*} The problem is, that I need $f(\gamma_i)$ instead of $f\left(\sum_{i = 1}^n \gamma_i(t)\right)$ on the right hand side. How can I achieve that?