<s>[INST] Show $\int \dfrac{\sinh x}{\cosh 2x}=\dfrac 1 {2\sqrt 2} \ln\left|\dfrac {\sqrt 2 \cosh x-1}{\sqrt 2 \cosh x +1}\right| + C$ 
Show by means of the substitution $u = \cosh x$, that
  $$\int \dfrac{\sinh x}{\cosh 2x}=\dfrac 1 {2\sqrt 2} \ln\left|\dfrac {\sqrt 2 \cosh x-1}{\sqrt 2 \cosh x +1}\right| + C$$

$$\int \dfrac{\sinh x}{\cosh 2x}= \dfrac {\sqrt 2}{2} \tanh ^{-1} (\sqrt 2) u + C$$
Let $$\tanh ^{-1} (\sqrt 2u) = y$$
Then $$2y = \ln \left| - \dfrac {\sqrt 2 u +1}{\sqrt 2 u -1}\right|$$
Given $\sqrt 2 u -1 \geq 0$ or since $u = \cosh x \geq 1$
$$y = \dfrac 1 2 \ln \left|\dfrac {\sqrt 2 u +1}{\sqrt 2 u -1}\right|$$
Giving
$$\int \dfrac{\sinh x}{\cosh 2x}=\dfrac 1 {2\sqrt 2} \ln\left|\dfrac {\sqrt 2 \cosh x+1}{\sqrt 2 \cosh x -1}\right| + C$$
Which is different to the given solution.
Where did I go wrong?
Many thanks in advance
 [/INST] I believe your mistake is missing the negative sign here:
$$\int \dfrac{\sinh x}{\cosh 2x}= -\dfrac {\sqrt 2}{2} \tanh ^{-1} (\sqrt 2) u + C$$
In lots more steps:
$$\begin{split}
\int \dfrac{\sinh x}{\cosh 2x}dx &= \int \dfrac{\sinh x}{2\cosh^2(x) - 1}dx \\
&= \int \dfrac{1}{2u^2 - 1}du \\
&= \dfrac{1}{\sqrt2}\int \dfrac{1}{v^2 - 1}dv \\
&= -\dfrac{1}{\sqrt2}\int \dfrac{1}{1 - v^2}dv (*) \\
&= -\dfrac{1}{\sqrt2}\tanh^{-1} v + C \\
&= -\dfrac{1}{\sqrt2}\tanh^{-1}(u\sqrt{2}) + C
\end{split}$$
Need to invert in $(*)$ because $\dfrac{d}{dx}\tanh^{-1}x = \dfrac{1}{1 - x^2}$. The rest follows from the missing negative.
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