We know for$ f(x) = \arctan(x)$, $f'(x) = \dfrac1{x^2 + 1}$. Now we can write $x^2 + 1$ as $(x - i)\cdot(x + i)$ where $i = \sqrt{-1}$. So we can express $\dfrac1{x^2 + 1}$ as $\lambda\cdot\left(\dfrac1{x - i} - \dfrac1{x + i}\right)$ where $\lambda=\dfrac1{2i}$.
$$\begin{align} \implies\arctan(x) &= \int\lambda\cdot\left(\dfrac1{x - i} - \dfrac1{x + i}\right)\mathrm dx \\ \implies\arctan(x) &= \lambda\cdot\ln\left|\dfrac{x - i}{x + i}\right| + c \end{align}$$ For $x=0$, we find that $c=0$. $$\implies\arctan(x)=\lambda\cdot\ln\left|\dfrac{x - i}{x + i}\right|$$ for all values of $x$ where both functions are defined. But according to wolfram alpha, the graphs of the two functions don't match so what am I doing wrong here?
This is incorrect as the domains on either side of the equality are different. If you assume that $x$ is real, how could you define $x-i$?
It would be OK if you wrote $z$ instead of $x$; taken as a variable in $\mathbb{C}$. Now try plotting both in W|A under this domain.