Does $\int \frac{1}{x+i}=ln|x+i|$, or is there something peculiar about using complex numbers?

Question

amistre64 · Answer

i spose i shoulda included a "dx" :)

anonymous · Answer

That should be right, any integral of 1/# will be natural log with or without complex numbers. As long as there isn't any multiplying or anything.

anonymous · Answer

i suppose no ... it is just treated as constant

amistre64 · Answer

if we decompose a fraction into a complex linear; how would that affect the outcome of the integration?

amistre64 · Answer

$$\int\frac{1}{x^2+1}dx=tan^{-1}(x)+C$$
but what if we decompose it?

zarkon · Answer

you could also do this...

$$\int\frac{1}{x+i}dx=\int\frac{1}{x+i}\frac{x-i}{x-i}dx$$
$$=\int\frac{x-i}{x^2+1}dx=\ln(x^2+1)-\tan^{-1}(x)i+c$$

zarkon · Answer

oops

$$=\int\frac{x-i}{x^2+1}dx=\frac{\ln(x^2+1)}{2}-\tan^{-1}(x)i+c$$

anonymous · Answer

@zarkon : its$$\frac{\ln(x ^{2}+1)}{2} - i  (\tan^{-1} (x ))+c$$

zarkon · Answer

I know