$\Gamma$ is a circle of radius 2 around the point $1+i$. I've parametrized the circle as $\gamma(t)=2e^{it}+1+i$ substituting $z$ in te integral for that expression gets really ugly really quickly. I'm assuming there is some simplification I'm missing that would let me compute the integral more easily.
I'm new to complex integration so I don't really know what I can and can't do. Any help would be appreciated.
This is the denominator after substitution. $2 ((4+8 i) e^{i t}+(6+8 i) e^{2 i t}+4 e^{3 i t}+(1+4 i))$
Using Residue theorem one has : $$\oint_\Gamma \dfrac{\ln(z+5)}{z^3 + iz^2 + 6z} dz = 2\pi i \sum_k Res(f, z_k),$$ where $f(z) = \dfrac{\ln(z+5)}{z^3 + iz^2 + 6z}$ and $z_k$ is some pole of $f$ inside $\Gamma$. As $\ln(z+5)$ has no pole inside $\Gamma$ one just focuses on poles of $\dfrac{1}{z^3 + iz^2 +6z}$. As $z^3 +iz^2 +6z = z (z-2i)(z+3i)$ the poles of $f$ inside $\Gamma$ are : $$z_1 = 0,\quad z_2 = 2i$$ To compute the residues one could use Laurent's series or the following fomula : $$Res(f,z_k) = \dfrac{1}{(n-1)!}\lim_{z\rightarrow z_k}\dfrac{\partial^{n-1}}{\partial z^{n-1}} (z-z_k)^n f(z)$$ $f$ has only simple poles i.e. $n=1$. Thus one finds : $$Res(f,z_1) = \lim_{z\rightarrow z_1} (z-z_1)^n f(z) = \dfrac{\ln(5)}{6},\quad Res(f,z_2) = \lim_{z\rightarrow z_2} (z-z_2)^n f(z) = -\dfrac{\ln(5+2i)}{10}$$ Hence $$\oint_\Gamma \dfrac{\ln(z+5)}{z^3 + iz^2 + 6z} dz =\pi i \left ( \dfrac{\ln(5)}{3} - \dfrac{\ln(5+2i)}{5}\right)$$