Answer
$$d\left( t \right) = \frac{{{v_T}}}{a}\ln \left( {\frac{{{e^{at}} + 2 + {e^{ - at}}}}{4}} \right)$$
Work Step by Step
$$\eqalign{
& {\text{Let }}v\left( t \right) = {v_T}\left( {\frac{{{e^{at}} - 1}}{{{e^{at}} + 1}}} \right) \cr
& {\text{Let }}y = t \cr
& v\left( y \right) = {v_T}\left( {\frac{{{e^{ay}} - 1}}{{{e^{ay}} + 1}}} \right) \cr
& {\text{Where }}d\left( t \right) = \int_0^t {v\left( y \right)} dy,{\text{ so}} \cr
& d\left( t \right) = \int_0^t {{v_T}\left( {\frac{{{e^{ay}} - 1}}{{{e^{ay}} + 1}}} \right)} dy \cr
& d\left( t \right) = {v_T}\int_0^t {\left( {\frac{{{e^{ay}} - 1}}{{{e^{ay}} + 1}}} \right)} dy \cr
& {\text{Distribute the numerator}} \cr
& d\left( t \right) = {v_T}\int_0^t {\left( {\frac{{{e^{ay}}}}{{{e^{ay}} + 1}}} \right)} dy - {v_T}\int_0^t {\left( {\frac{1}{{{e^{ay}} + 1}}} \right)} dy \cr
& d\left( t \right) = \frac{{{v_T}}}{a}\int_0^t {\left( {\frac{{a{e^{ay}}}}{{{e^{ay}} + 1}}} \right)} dy + \frac{{{v_T}}}{a}\int_0^t {\left( {\frac{{ - a{e^{ - ay}}}}{{1 + {e^{ - ay}}}}} \right)} dy \cr
& {\text{Integrating}} \cr
& d\left( t \right) = \frac{{{v_T}}}{a}\left[ {\ln \left( {{e^{ay}} + 1} \right)} \right]_0^t + \frac{{{v_T}}}{a}\left[ {\ln \left( {1 + {e^{ - ay}}} \right)} \right]_0^t \cr
& d\left( t \right) = \frac{{{v_T}}}{a}\left[ {\ln \left( {{e^{ay}} + 1} \right) + \ln \left( {1 + {e^{ - ay}}} \right)} \right]_0^t \cr
& d\left( t \right) = \frac{{{v_T}}}{a}\left[ {\ln \left[ {\left( {{e^{ay}} + 1} \right)\left( {1 + {e^{ - ay}}} \right)} \right]} \right]_0^t \cr
& {\text{Evaluating}} \cr
& d\left( t \right) = \frac{{{v_T}}}{a}\left[ {\ln \left( {{e^{at}} + 1} \right)\left( {1 + {e^{ - at}}} \right) - \ln \left( {{e^0} + 1} \right)\left( {1 + {e^0}} \right)} \right] \cr
& d\left( t \right) = \frac{{{v_T}}}{a}\left[ {\ln \left( {{e^{at}} + 1 + 1 + {e^{ - at}}} \right) - \ln 4} \right] \cr
& d\left( t \right) = \frac{{{v_T}}}{a}\left[ {\ln \left( {{e^{at}} + 2 + {e^{ - at}}} \right) - \ln 4} \right] \cr
& d\left( t \right) = \frac{{{v_T}}}{a}\ln \left( {\frac{{{e^{at}} + 2 + {e^{ - at}}}}{4}} \right) \cr} $$
