Answer
$$\ln \left( {\cosh x} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\tanh x} dx \cr
& {\text{hyperbolic identity tanh}}\phi = \frac{{\sinh \phi }}{{\cosh \phi }} \cr
& = \int {\frac{{\sinh x}}{{\cosh x}}} dx \cr
& {\text{substitute }}u = \cosh x,{\text{ }}du = \sinh xdx \cr
& = \int {\frac{{\sinh x}}{{\cosh x}}} dx = \int {\frac{{du}}{u}} \cr
& = \int {\frac{{du}}{u}} \cr
& {\text{find the antiderivative}} \cr
& = \ln \left| u \right| + C \cr
& {\text{write in terms of }}x \cr
& = \ln \left| {\cosh x} \right| + C \cr
& \cosh x{\text{ is positive for all x}} \cr
& = \ln \left( {\cosh x} \right) + C \cr} $$
