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$\DeclareMathOperator{\arcsinh}{arcsinh}$I have seen that $$ \arcsinh(x) = \ln(x + \sqrt{x^2 + 1}) \tag{1} $$ and also that $$ \arcsinh(x/a) = \ln(x + \sqrt{x^2 + a^2}). \tag{2} $$

I have to calculate $\arcsinh(\frac{1}{2}\sqrt{2})$; the above formula give values of $\frac{1}{2} \ln 2$ and $\ln 4$, respectively.

My question is what have i done wrong/assumed. I know for certain that both formula are correct as they are in official formula booklet.

Andrew D. Hwang
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Robert S
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    If you replace $x$ with $x/a$ in the first formula (assuming $x$ and $a$ are positive), you get $$\ln(x+\sqrt{x^2+a^2})-\ln a$$ which does not match your second formula. – MPW May 13 '16 at 21:37
  • Yes, but nonetheless both formula are correct. The problem is that clearly one should be used under certain circumstances while the other is used at different times. I'm seeking clarification as to what those conditions are – Robert S May 13 '16 at 21:39
  • I am thinking that it is to do with x being a variable while a is a constant but am unable to justify the difference yet – Robert S May 13 '16 at 21:41
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    Is your booklet a table of integrals/list of antiderivative formulas, by chance? (Incidentally, the first formula is algebraically correct for all real $x$. As MPW notes, the second is therefore missing an additive $-\ln a$.) – Andrew D. Hwang May 13 '16 at 22:04
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    the second is part of a list of integrals – Robert S May 13 '16 at 23:03

1 Answers1

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The second formula is incorrect. You are missing a term "$-\ln a$" which should be appended to the end.

If you have truly quoted it verbatim from a booklet, then the booklet is wrong.

Is it possible that the extra term was printed on a second line and you simply missed it?

As @Andrew D. Hwang hints, if these are antiderivatives, they should both have a "$+C$" tacked on, and that could absorb the missing constant; but as written, he formula is wrong.

MPW
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