Prove 2coth^(-1)(e^x) = ln[cosech x + coth x]

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Homework Statement
Prove 2coth^(-1)(e^x) = ln[cosech x + coth x]

The attempt at a solution
I'm starting with the L.H.S. first, which is conventional, i think.

2coth^(-1)(e^x)

= 2tanh^(-1)(e^-x)

= 2[1/2 ln [(1+(e^-x))/(1-(e^-x))]

= ln [(1+(e^-x))/(1-(e^-x))]

I'm stuck.

So, i tried to cheat a little, by expanding the R.H.S. and see if i could find a way to convert the last step above into the R.H.S. but my copybook has become something of a painting.
 
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sharks said:
Homework Statement
Prove 2coth^(-1)(e^x) = ln[cosech x + coth x]

The attempt at a solution
I'm starting with the L.H.S. first, which is conventional, i think.

2coth^(-1)(e^x)

= 2tanh^(-1)(e^-x)

= 2[1/2 ln [(1+(e^-x))/(1-(e^-x))]

= ln [(1+(e^-x))/(1-(e^-x))]

I'm stuck.

So, i tried to cheat a little, by expanding the R.H.S. and see if i could find a way to convert the last step above into the R.H.S. but my copybook has become something of a painting.

Hints: work from the RHS, and substitute u = e^x. Will make everything much clearer.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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