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$$ \int \frac{dx}{1+x^2} = \arctan x + C $$

This is achieved with a basic trig substitution. However, what if one were to perform the following partial fraction decomposition:

$$ \int \frac{dx}{1+x^2} = \int \frac{dx}{(x+i)(x-i)} = \int \left[ \frac{i/2}{x+i} - \frac{i/2}{x-i} \right] \, dx $$

I believe that this is justified because plugging in any real number ##a## into the integrand would simplify to a real number (namely ##1+a^2##).

Thus,

$$ \int \frac{dx}{1+x^2} = (i/2) \ln |x+i| - (i/2) \ln |x-i| + C $$

In conclusion:

$$ \arctan z = \frac{i}{2} \ln \left| \frac{z+i}{z-i} \right| $$

The right side should yield real value outputs for real value inputs.

Word?

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# Partial Fractions with Complex Numbers

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