Evaluate $$\int_C \frac{1}{1+z^2} dz$$ along the line segment from $z=1$ to $z=1+i$.
My solution:
Parametrization $p(t) = 1+t, \:\: 0\leq t \leq i$ with derivative $1$. We thus get: $$\int_C \frac{1}{1+z^2} dz = \int_{0}^{i} \frac{1}{1+(1+t)^2} dt = tan^{-1}(1+t)|_{0}^{i} = tan^{-1}(1+i)-tan^{-1}(1)$$
This is however wrong. So my question is, where did I go wrong?
Hint: the parametrisation is something that you can improve.
Set
$$z(t) = 1 + it ~~~~~~~ t\in [0, 1]$$
$$z'(t) = i$$
Hence use the formula:
$$\int_{\gamma_1}f(z)\ \text{d}z = \int_{\gamma}(f(z(t))z'(t)\ \text{d}t = \int_0^1 \frac{i\ \text{d}t}{1 + (1+it)^2}$$
The denominator is not:
$$1 + (1+it)^2 = 1 + 1 + 2it - t^2 = -t^2 + 2it + 2$$
Poles are at
$$t^2 - 2it - 2 = 0 ~~~~~~~ \to t = \begin{cases} i + 1 \\ i-1 \end{cases}$$
Hence the integral can be written as
$$i\int_0^1 \frac{\text{d}t}{(t-i-1)(t-i+1)}$$
Now you should be able to evaluate it with normal integration techniques.
Can you proceed?
Final Solution
By splitting the integral as you did in the comment below, you will get two really trivial integrals:
$$\frac{i}{2}\int_0^1 \frac{\text{d}t}{t - (i+1)} = \frac{1}{8}\left(-\pi - i\ln(4)\right)$$
and
$$\frac{i}{2}\int_0^1 \frac{\text{d}t}{t - (i-1))} = \frac{1}{8} \left(-\pi +i \log \left(\frac{25}{4}\right)+4 \tan ^{-1}\left(\frac{1}{2}\right)\right)$$
You really can obtain those results in the trivial way in which you obtained your results in what you tried.
Now, the final result is the difference:
$$\frac{1}{8}\left(-\pi - i\ln(4)\right) - \frac{1}{8} \left(-\pi +i \log \left(\frac{25}{4}\right)+4 \tan ^{-1}\left(\frac{1}{2}\right)\right) = \frac{1}{8}\left(-i\ln(25) + 4\arctan\left(\frac{1}{2}\right)\right)$$
Notice that you can funny manipulate. For example: $25 = 5^2 \to \ln(25) = 2\ln(5)$ et cetera.