Answer
$7 \ln ( \cosh \dfrac{x}{7}) +C= 7 \ln |e^{x/7}+e^{-x/7}| +C$
Work Step by Step
Given: $\int \tanh (\dfrac{x}{7}) dx$
Plug in $\dfrac{x}{7}=a$ and $dx= 7 da$
Thus, $\int \tanh (\dfrac{x}{7}) dx= 7 \int \tanh a da =7 \int \dfrac{\sinh a}{\cosh a} da $
Now, plug in $\cosh a =u \implies \sinh a da =du$
Thus, we have:
$=7 \int \dfrac{dt}{t}$
or, $= 7 \ln |u| +C$
or, $= 7 \ln |\cosh \dfrac{x}{7}| +C$
$= 7 \ln |\frac{e^{x/7}+e^{-x/7}}{2}| +C$
$= 7 \ln |e^{x/7}+e^{-x/7}| +C$
