at school, we solved this integral and the solution we got was

enter image description here

Wolfram Alpha gave a different solution

Integral solution 2

Are these two solutions equal?


Yes, they are. You can work out this through very simple algebraic means. One has

$$\tanh x=\frac{e^x-e^{-x}}{e^{x}+e^{-x}}.$$

Now $z=\tanh x$, put $y=x$ and you will get the following


and solve for $y$. You will get


and so


and $\tanh^{-1}$ in this case is obtained taking the logarithm and keeping just the positive solution, that is,


So, if you put $z=\sqrt{1+e^{2x}}$ you will get the identity.


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