Integration of $\log(2\cos(x)e+e^2+1)$

Question

Solve integral $$\int_\gamma{\frac{\log(z+e)}{z}}dz,\hspace{3mm} \text{where} \hspace{3mm}\gamma(t)=e^{it},\hspace{3mm} t\in[0,2\pi]$$

I wanted to solve this without Cauchy's integral formula and I already got

$$ \int_\gamma{\frac{\log(z+e)}{z}}dz=i\int_0^{2\pi}{\frac{\log(e^{it}+e)e^{it}}{e^{it}}}dt \\ =i\int_0^{2\pi}{\log\sqrt{|2e\cos{t}+e^2+1|}+i\int_0^{2\pi}\arctan\left(\frac{\sin{t}}{\cos{t}+e}\right)} \\ =\frac{i}{2}\int_0^{2\pi}\log(2e\cos{t}+e^2+1) $$

Wolfram gives the result $2\pi i$, so it looks neat. It is possible to solve integral $\int_0^{\frac{\pi}{2}}{\log(\cos{t})}$ so I thought I could solve this integral too. I have no clue how to solve this if it is possible.

Links:

Evaluate $\int_0^{\pi/2}\log\cos(x)\,\mathrm{d}x$

Answer 1

See this answer. $\cos{(2\pi-x)}=\cos{x}$, so your last integral is double the considered one. Because $e>1$, this then gives $4\pi \log{e} = 4\pi$ as the value, and you find Wolfram|Alpha's answer.