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Right. In my textbook, it says:

arsinh(x) = u

...

=> e^{u} = x +/- √(x^{2} + 1)

But we ignore the negative root since:

√(x^{2} + 1) > x

=> e^{u} < 0 , which is not possible

giving arsinh x = ln[x + √(x^{2} + 1)]

However, this is not the case for arcosh

arcosh(x) = u

...

=> e^{u} = x +/- √(x^{2} - 1)

but √(x^{2} - 1) is not necessarily more than x

but the identity given for arcosh is:

arcosh = ln[x + √(x^{2} - 1)]

so, if we exclude the -ve root, will we not lose some answers?

arsinh(x) = u

...

=> e

But we ignore the negative root since:

√(x

=> e

giving arsinh x = ln[x + √(x

However, this is not the case for arcosh

arcosh(x) = u

...

=> e

but √(x

but the identity given for arcosh is:

arcosh = ln[x + √(x

so, if we exclude the -ve root, will we not lose some answers?

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