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We can express $\tanh(\ln(1+x^2))$ as a sum of its real and imaginary parts using the following identities:

$$\begin{align*} \tanh(x) &= \frac{\sinh(x)}{\cosh(x)}, \sin(x) &= \frac{1}{2i}(e^{ix}-e^{-ix}), \cos(x) &= \frac{1}{2}(e^{ix}+e^{-ix}). \end{align*}$$

Substituting $x = \ln(1+x^2)$, we get:

$$\begin{align*} \tanh(\ln(1+x^2)) &= \frac{\sinh(\ln(1+x^2))}{\cosh(\ln(1+x^2))} &= \frac{\sinh(\ln(1+x^2))}{\sqrt{\cosh^2(\ln(1+x^2))}} &= \frac{\sinh(\ln(1+x^2))}{\sqrt{\sinh^2(\ln(1+x^2))+1}} &= \frac{\frac{1}{2}(e^{\ln(1+x^2)}-e^{-\ln(1+x^2)})}{\sqrt{\frac{1}{4}(e^{2\ln(1+x^2)}+2+e^{-2\ln(1+x^2)})}} &= \frac{1}{2i}\cdot\frac{e^{i\ln(1+x^2)}-e^{-i\ln(1+x^2)}}{\sqrt{e^{2i\ln(1+x^2)}+2+e^{-2i\ln(1+x^2)}}} &= \frac{1}{2i}\cdot\frac{(1+x^2)^i-(1+x^2)^{-i}}{\sqrt{(1+x^2)^2+1-2(1+x^2)\cos(2\ln(1+x^2))}} &= \frac{1}{2i}\cdot\frac{(1+x^2)^i-(1+x^2)^{-i}}{\sqrt{(1+x^2)^2-2(1+x^2)+2}} &= \frac{1}{2i}\cdot\frac{(1+x^2)^i-(1+x^2)^{-i}}{\sqrt{x^4+2x^2+1}}. \end{align*}$$

Therefore, we have expressed $\tanh(\ln(1+x^2))$ as a sum of its real and imaginary parts:

$$\tanh(\ln(1+x^2)) = \frac{x^2}{x^2+1}+\frac{1}{2i}\cdot\frac{(1+x^2)^i-(1+x^2)^{-i}}{\sqrt{x^4+2x^2+1}}.$$