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We can write the real part of $\tanh(\ln(1+x^2))$ in terms of exponential functions using the identity $\tanh(x) = \frac{e^{2x}-1}{e^{2x}+1}$. First, we have $$\tanh(\ln(1+x^2)) = \frac{\sinh(\ln(1+x^2)))}{\cosh(\ln(1+x^2))+1} = \frac{x}{\sqrt{1+x^2}+1},$$ where we have used the hyperbolic sine and cosine identities $\sinh(\ln(1+x^2)) = \frac{x}{2}(1+x^2)^{\frac{1}{2}}$ and $\cosh(\ln(1+x^2)) = \frac{1+x^2}{2}$.

Using the identity $\tanh(x) = \frac{e^{2x}-1}{e^{2x}+1}$, we can write $$\tanh(\ln(1+x^2)) = \frac{e^{2\ln(1+x^2)}-1}{e^{2\ln(1+x^2)}+1} = \frac{x^2}{x^2+1}.$$

Therefore, we have two different closed-form expressions for the real part of $\tanh(\ln(1+x^2))$, namely $$\operatorname{Re}(\tanh(\ln(1+x^2))) = \frac{x}{\sqrt{1+x^2}+1} \quad\text{and}\quad \operatorname{Re}(\tanh(\ln(1+x^2))) = \frac{x^2}{x^2+1}.$$