Complex integration: $\int _\gamma \frac{1}{z}dz=\log (\gamma (b))-\log(\gamma (a))?$ Let $\gamma$ be a closed path defined on $[a,b]$ with image in the complex plan except the upper imaginary axis, ($0$ isn't in this set).
Then $\frac{1}{z}$ has an antiderivative there and it is $\log z$. Therefore $\int _\gamma \frac{1}{z}dz=\log (\gamma (b))-\log(\gamma (a))=0$ because it is a closed path.
Now let $\psi(t)=e^{it}+3$, $0\leq t\leq 2\pi$. Then $\psi'(t)=ie^{it}, 0\leq t\leq 2\pi$.
So $$\int _\psi\frac{1}{z}dz=\int _0^{2\pi}\frac{ie^{it}}{e^{it}}dt=2\pi i$$
but $\psi$ is a closed path so there's something wrong.
What's going on here?

The denominator in the second expression should be $e^{it}+3$ instead of $e^{it}$.
