Answer
$$\frac{1}{5}$$
Work Step by Step
$$\eqalign{
& \int_{\ln \left( {3/4} \right)}^{\ln \left( {4/3} \right)} {\frac{{{e^t}dt}}{{{{\left( {1 + {e^{2t}}} \right)}^{3/2}}}}} \cr
& {\text{We set}}{\text{, }}{e^t} = \tan \theta,{\text{ }}{e^t}dt = {\sec ^2}\theta d\theta \cr
& \theta = {\tan ^{ - 1}}\left( {{e^t}} \right) \cr
& {\text{The new limits on }}\theta {\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}t = \ln \left( {4/3} \right),{\text{ then }}\theta = {\tan ^{ - 1}}\left( {{e^{\ln 4/3}}} \right) = {\tan ^{ - 1}}\left( {4/3} \right) \cr
& \,\,\,\,\,\,{\text{If }}t = \ln \left( {3/4} \right),{\text{ then }}\theta = {\tan ^{ - 1}}\left( {{e^{\ln 3/4}}} \right) = {\tan ^{ - 1}}\left( {3/4} \right) \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& \int_{\ln \left( {3/4} \right)}^{\ln \left( {4/3} \right)} {\frac{{{e^t}dt}}{{{{\left( {1 + {e^{2t}}} \right)}^{3/2}}}}} = \int_{{{\tan }^{ - 1}}\left( {3/4} \right)}^{{{\tan }^{ - 1}}\left( {4/3} \right)} {\frac{{{{\sec }^2}\theta d\theta }}{{{{\left( {1 + {{\tan }^2}\theta } \right)}^{3/2}}}}} \cr
& {\text{use the fundamental identity }}{\tan ^2}x + 1 = {\sec ^2}x \cr
& = \int_{{{\tan }^{ - 1}}\left( {3/4} \right)}^{{{\tan }^{ - 1}}\left( {4/3} \right)} {\frac{{{{\sec }^2}\theta d\theta }}{{{{\left( {{{\sec }^2}\theta } \right)}^{3/2}}}}} \cr
& = \int_{{{\tan }^{ - 1}}\left( {3/4} \right)}^{{{\tan }^{ - 1}}\left( {4/3} \right)} {\frac{{{{\sec }^2}\theta d\theta }}{{{{\sec }^3}\theta }}} \cr
& = \int_{{{\tan }^{ - 1}}\left( {3/4} \right)}^{{{\tan }^{ - 1}}\left( {4/3} \right)} {\frac{{d\theta }}{{\sec \theta }}} \cr
& = \int_{{{\tan }^{ - 1}}\left( {3/4} \right)}^{{{\tan }^{ - 1}}\left( {4/3} \right)} {\cos \theta d\theta } \cr
& {\text{integrating}} \cr
& = \left( {\sin \theta } \right)_{{{\tan }^{ - 1}}\left( {3/4} \right)}^{{{\tan }^{ - 1}}\left( {4/3} \right)} \cr
& = \sin \left( {{{\tan }^{ - 1}}\left( {4/3} \right)} \right) - \sin \left( {{{\tan }^{ - 1}}\left( {3/4} \right)} \right) \cr
& = \frac{4}{5} - \frac{3}{5} \cr
& = \frac{1}{5} \cr} $$
