Inverse Hyperbolic Trigonometry as Logarithms: coth^-1(x)

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In this video I go over converting inverse hyperbolic trig functions to logarithms and prove that the function inverse hyperbolic cotangent or coth^-1(x) is equal to 1/2·ln((x+1)/(x-1)).

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I don't always write functions as logs but when I do it's usually when dealing with inverse hyperbolic trig ;)

mes
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I appreciate how you simplify and don't complicate it

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