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Show that $$f(x) = \ln \left(x+\sqrt{x^2+1}\right)$$ is an odd function.

My attempt:

$$f(-x)=\ln\left(-x+\sqrt{(-x)^2+1}\right)=\ln\left(-x+\sqrt{x^2+1}\right).$$

How should I proceed? I know that if $f(-x)=-f(x)$, the function is odd.

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    $\begingroup$ "I know that et cetera" ... so you must show, and it suffices showing, that $-x+\sqrt{x^2+1}=\frac{1}{x+\sqrt{x^2+1}}$ for all $x$. $\endgroup$ – user228113 Aug 15 '17 at 16:30
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Try to use $$(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)=1$$

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You are almost done : Simply note that $f(x)+ f(-x) = \ln(\sqrt {x^2 + 1} + x) + \ln(\sqrt{x^2 + 1}-x) = \\ \ln((\sqrt {x^2 + 1} + x)(\sqrt{x^2 + 1}-x)) = \ln (x^2 + 1 - x^2) = \ln 1 = 0$.

Hence, the function is an odd function.

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$$\ln(-x+\sqrt{x^2+1})=\ln\left(\frac{1}{\sqrt{x^2+1}+x}\right)=-\ln(\sqrt{x^2+1}+x)$$

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  • $\begingroup$ How come you could take its reciprocal? $\endgroup$ – Andrew Tawfeek Aug 15 '17 at 17:57
  • $\begingroup$ I am just multiplying on the top and bottom by $\sqrt{x^2+1}+x$ $\endgroup$ – tattwamasi amrutam Aug 15 '17 at 18:41
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Note that $\sqrt{1+x^2}$ is even function.

Now, $$\int_{-x}^{x}\frac{1}{\sqrt{1+t^2}}dt=2\int_{0}^{x}\frac{1}{\sqrt{1+t^2}}dt=2\ln(x+\sqrt{1+x^2})$$

Again, $$\int_{-x}^{x}\frac{1}{\sqrt{1+t^2}}dt=2\int_{-x}^{0}\frac{1}{\sqrt{1+t^2}}dt=-2\ln(-x+\sqrt{1+x^2})$$

Hence $\ln(x+\sqrt{1+x^2})=\ln(-x+\sqrt{1+x^2})$ i.e. odd function.

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$$ f(\sinh\theta) = \theta $$ and both $\theta$ and $\sinh\theta$ are odd functions, hence $f$ is an odd function as well ($f(x)=\text{arcsinh}(x)$).

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One more:

$f(x) = \ln (x + \sqrt{x^2+1})$, $x \in \mathbb{R}.$

$\star)$ $f(-x) = \ln(-x + \sqrt{x^2 +1}) =$

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

$- \ln (x + \sqrt{x^2+1})$.

Combining:

$f(-x) = - f(x)$.

Used :

$ \ln [( -x +\sqrt{x^2+1}) \frac{x + \sqrt{x^2 +1}}{x+\sqrt{x^2+1}}] =$

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

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