Answer
$\dfrac{1}{2}\ln (\dfrac{17}{8})$
Work Step by Step
The given integral $\int^{\ln 2}_{0} \tanh 2x dx$ can be re-written as:$\int^{\ln 2}_{0} \dfrac{\sinh 2x}{\cosh 2x} dx$
Now, consider $\cosh 2x=t$ and $dt=2 \sinh 2x dx$
Now, $\int^{\ln 2}_{0} \dfrac{\sinh 2x}{\cosh 2x} dx=\dfrac{1}{2}\int^{17/8}_{1} (\dfrac{dt}{t})=\dfrac{1}{2}[\ln |t|]^{17/8}_{1}$
or, $=\dfrac{1}{2}[\ln (\dfrac{17}{8})-\ln(1)] $
So, $=\dfrac{1}{2}\ln (\dfrac{17}{8})$
