Complex

\log (1+\frac{i}{x})=\frac{i}{x}-\frac{i^2}{2x^2}+\frac{i^3}{3x^3}-\cdots

i\log (x-i)-i\log (x+i)=i\log (1-\frac{i}{x})-i\log (1+\frac{i}{x})
=-\frac{i^2}{x}-\frac{i^3}{2x^2}-\frac{i^4}{3x^3}-\cdots-\frac{i^2}{x}+\frac{i^3}{2x^2}-\frac{i^4}{3x^3}+\cdots
=2\arctan\frac{1}{x}=\pi-2\arctan x

i\log (x+i)+i^2+i^3\log (x-i)+i^4(2\arctan x)=\pi-1

Another way of doing the problem ,

e^{i\arctan \frac{1}{x}}=\frac{x}{\sqrt{x^2+1}}+\frac{i}{\sqrt{x^2+1}}

Now take log on both sides to find log(x+i).Do the same for log(x-i).

Are you sure about the question(or the answer)?Substitute x=0 and you'll find that the answer is π-1.