Consider
$$\int_\gamma \frac{1}{4z+iz^2}\mbox{d}z $$
where $\gamma$ is the boundary of the triangle with vertices $-i,2+i,-2+i$.
The assignment is to solve it using the definition which is given as
$$\int_\gamma f(z)\mbox{d}z \overset{def}= \int_\alpha^\beta f(z(t))z'(t)\mbox{d}t $$
Where we have parametrized the curve $\gamma$ and $\alpha\leq t\leq\beta$.
I have parametrized the boundary in three parts
$$\gamma _1: \begin{cases}x(t) = 2t\\y(t)=-1+2t\end{cases},\,0\leq t\leq 1\bigg|\gamma _2: \begin{cases}x(t) = 2-4t\\y(t)=1\end{cases},\,0\leq t\leq 1\\
\gamma _3: \begin{cases}x(t) =-2+2t\\y(t)=1-2t\end{cases},\,0\leq t\leq 1 $$
Since $\gamma _1\cup\gamma _2\cup\gamma _3 =\gamma$ we have:
$$\int_\gamma f(z)\mbox{d}z = \int_{\gamma _1} f(z)\mbox{d}z+\int_{\gamma _2} f(z)\mbox{d}z+\int_{\gamma _3} f(z)\mbox{d}z $$
BUT computing these integrals seems nigh impossible, an example: $$\int_{\gamma _1}\frac{1}{4z+iz^2}\mbox{d}z = \int_0^1 \frac{(2t)' +i(-1+2t)'}{4(2t+i(-1+2t))+i(2t+i(-1+2t))^2}\mbox{d}t $$
How to proceed to solve the problem using Only the definition? (Newton-Leibniz and Cauchy are out of the question for now)
With a little work, you can simplify the integral you have posted (which I verified to be correct). Multiplying everything out, I get
$$(1+i) \int_0^1 \frac{dt}{-4 t^2 + 9 (1+i) t-i 10} $$
The roots of the denominator are at $t=1+i$ and $t=\frac54(1+i)$. The integral is then
$$(1+i) \int_0^1 \frac{dt}{(t-\frac54 (1+i))(t-(1+i))} = 4 \int_0^1 dt \left (\frac1{t-\frac54(1+i)} - \frac1{t-(1+i)} \right ) $$
This one you should be able to do; the result is $4 \log{(1-i/5)}$.
I imagine the other ones work out as nicely.