Originally Posted by

**Soroban** Hello, wonderboy1953!

TheCoffeeMachine is right . . . The left side needs an $\displaystyle \,i.$

I derived this, too, many many years ago.

Here's how I did it.

$\displaystyle \displaystyle \text{We know that: }\:\int\frac{dx}{x^2+1} \;=\;\tan^{-1}x + C_1$ .[1]

$\displaystyle \text{Since }\:x^2+1 \;=\;x^2 - (\text{-}1) \;=\;x^2 - i^2 \;=\;(x-i)(x+i)$

. . $\displaystyle \displaystyle \text{then: }\;\int\frac{dx}{x^2+1} \;=\;\int\frac{dx}{(x-i)(x+i)} \;=\;\frac{1}{2i}\ln\left|\frac{x-i}{x+i}\right| + C_2$ .[2]

Equate [1] and [2]: . $\displaystyle \displaystyle \tan^{-1}x + C_1 \;=\;\frac{1}{2i}\ln\left|\frac{x-i}{x+i}\right| + C_2$

$\displaystyle \displaystyle\text{Therefore: }\;2i\tan^{-1}x \;=\;\ln\left|\frac{x-i}{x+i}\right| + C$

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$\displaystyle \displaystyle \text{Another identity: }\;i\ln\left|\frac{x-1}{x+1}\right| \;=\;2\tan^{1}\left(\frac{x}{i}\right) + C$

$\displaystyle \displaystyle \text{This comes from: }\:\int\frac{dx}{x^2-1} \;=\;\int\frac{dx}{x^2+i^2}$