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Hyperbolic functions and logarithms

I need to show that:

(root5 - 2)ln(root5 - 2) + (root5 + 2)ln(root5 + 2) = 4 arsinh(2).

I have no idea. Any help would be appreciated.
Do you know the natural log equivalent of arsinh?

If so, depending on the number of marks, you'd be either expected to derive it and then plug in x=2, or just plug in x=2 if it's worth about 1 or 2 marks.
Original post by Protoxylic
Do you know the natural log equivalent of arsinh?

If so, depending on the number of marks, you'd be either expected to derive it and then plug in x=2, or just plug in x=2 if it's worth about 1 or 2 marks.


I know that arsinh x = ln(x+sqrt(x^2 + 1).

There's no number of marks; I'm working on an integral that I made up and I managed to solve it except that Mahtematica simplifies the expression I posted to 4 arsinh(2) and I can't see how.

Subbing x = 2 only gives 4 arsinh 2 = 4 ln(2+sqrt(5)) = ln(161+72root5), but the expression I posted isn't that, so there's still something I'm missing.
Simplify the ln expression and see what you get?


Posted from TSR Mobile
Original post by drandy76
Simplify the ln expression and see what you get?


Posted from TSR Mobile

Ok I got it now: the expression becomes root5(ln(root5-2)+ln(root5+2)-2(ln(root5-2)-ln(root5+2))
= root5*ln1 - 2ln((root5-2)^2) = -4ln(root5-2) = 4ln(1/(root5-2)) = 4ln(root5+2), as required.
Original post by HapaxOromenon3
I know that arsinh x = ln(x+sqrt(x^2 + 1).

There's no number of marks; I'm working on an integral that I made up and I managed to solve it except that Mahtematica simplifies the expression I posted to 4 arsinh(2) and I can't see how.

Subbing x = 2 only gives 4 arsinh 2 = 4 ln(2+sqrt(5)) = ln(161+72root5), but the expression I posted isn't that, so there's still something I'm missing.


You can simplify that, just notice that 1/[2+sqrt(5)]=sqrt(5)-2

Edit: You got it :tongue:

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