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So I am trying to simplify $\tanh(\operatorname{arsinh}(x))$ to $\frac{x}{\sqrt{1+x^2}}$

In general,

$$\tanh(x) = \frac{e^x-e^{-x}}{e^x+e^{-x}}$$

and $$\operatorname{arsinh}(x)= \ln(x+\sqrt{x^2-1})$$

therefore

\begin{align}\tanh(\operatorname{arsinh}(x)) & =\frac{e^{\ln(x+\sqrt{x^2-1})}-e^{\ln(\frac{1}{x+\sqrt{x^2-1}})}}{e^{\ln(x+\sqrt{x^2-1})}+e^{\ln(\frac{1}{x+\sqrt{x^2-1}})}}\\ &=\frac{x+\sqrt{x^2-1}-\frac{1}{x+\sqrt{x^2-1}}}{x+\sqrt{x^2-1}+\frac{1}{x+\sqrt{x^2-1}}} \\ &=\frac{(x+\sqrt{x^2-1})^2-1}{(x+\sqrt{x^2-1})^2+1} \\ &=\frac{x^2+2x\sqrt{x^2-1}+x^2-1-1}{x^2+2x\sqrt{x^2-1}+x^2-1+1} \\ &=\frac{x^2+x\sqrt{x^2-1}-1}{x^2+x\sqrt{x^2-1}} \\ &=\frac{x^2+x\sqrt{x^2-1}}{x^2+x\sqrt{x^2-1}}-\frac{1}{x^2+x\sqrt{x^2-1}} \\ &=1-\frac{1}{x(x+\sqrt{x^2-1})} \\ &=1-\frac{x-\sqrt{x^2-1}}{x(x+\sqrt{x^2-1})(x-\sqrt{x^2-1})} \\ &=1-\frac{x-\sqrt{x^2-1}}{x(x^2-x^2+1)} \\ &=1-\frac{x-\sqrt{x^2-1}}{x} \\ &=\frac{\sqrt{x^2-1}}{x} \\ &\ne \frac{x}{\sqrt{1+x^2}} \end{align}

I'm struggling to get my answer into the required form.

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  • $\begingroup$ $arsinh(x)=\ln{(x+\sqrt{x^2+1})}$ $\endgroup$ – AEngineer May 12 '18 at 9:41
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Your route is fine, but you have just made a mistake from the beginning, observe that $$ \operatorname{arsinh} x =\ln \left ( x + \sqrt{x^2 \color{red}{+} 1} \right )\ne \ln(x+\sqrt{x^2-1}). $$

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Let arcsinh$(x)=y\implies2x=2\sinh(y)=e^y-e^{-y}$

$\implies(e^y+e^{-y})^2=(e^y-e^{-y})^2+4e^y\cdot e^{-y}=4(x^2+1)$

and tanh$(y)=\dfrac{e^y-e^{-y}}{e^y+e^{-y}}=?$

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Let $$ arsinh(x)=y$$

$$ x= sinh (y)$$

$$ \sqrt {1+x^2} = cosh (y)$$

$$tanh ^2 (y) = 1- sech ^2 (y) = 1-\frac {1}{1+x^2}=\frac {x^2}{1+x^2} $$

$$tanh (y)=\frac {x}{\sqrt {1+x^2}}$$

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