Complex integration

I need help integrating $\int_{ |z+\frac{3i}{2}|} \frac{1}{z^2 +i} dz$

So I have split it int

$\frac{1}{4}\int_{\lambda} \frac{1}{2i} (\frac{1}{z-i} - \frac{1}{z+i})dz$.

Now I'm unsure on how to parametrise the circle. I've been told to use $z - i = e^{it}$ but don't know how to replace it for $z+i$.

Any help would be appreciated.

$|z+\frac{3}{2}i|$ is a circle centred at $z = -\frac{3}{2}i$, but the radius is unclear. Are you sure you've written this correctly?

Also $\dfrac{1}{z^2+i} \neq \dfrac{1}{2i} \left( \dfrac{1}{z-i} - \dfrac{1}{z+i}\right)$

Forgot to mention the radius is r. I've tried splitting it into partial fractions.

So the circle you are integrating along is $z = \frac{3}{2}i + Re^{i \theta}$ where $\theta$ ranges from 0 to $2\pi$.

You split that into partial fraction unsuccessfully... try again, or post your working out. Did you mean the fraction to be $\dfrac{1}{z^2 + 1}$ instead?

I found that the denominator $z^2 + i = (z+i)(z-i)$ and then worked from there. I've decided to abandoned this method of mine, and would use your first line and the original fraction instead, since I got quite confused where I am integrating along.

Thank you