## Calculus with Applications (10th Edition)

$$\frac{1}{8} - \frac{1}{{2\left( {3 + {e^2}} \right)}}$$
\eqalign{ & \int_0^1 {\frac{{{e^{2t}}}}{{{{\left( {3 + {e^{2t}}} \right)}^2}}}} dt \cr & {\text{use substitution}}{\text{. Let }}u = 3 + {e^{2t}},{\text{ so that }}du = 2{e^{2t}},\,\,\,\,\,\,\,\frac{1}{2}du = {e^{2t}}dt \cr & {\text{the new limits on }}u{\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}x = 1,{\text{ then }}u = 3 + {e^{2\left( 1 \right)}} = 3 + {e^2} \cr & \,\,\,\,\,\,{\text{If }}x = 0,{\text{ then }}u = 3 + {e^{2\left( 0 \right)}} = 4 \cr & {\text{Then}} \cr & \int_0^1 {\frac{1}{{{{\left( {3 + {e^{2t}}} \right)}^2}}}{e^{2t}}} dt = \int_4^{3 + {e^2}} {\frac{1}{{{u^2}}}\left( {\frac{1}{2}du} \right)} \cr & = \frac{1}{2}\int_4^{3 + {e^2}} {{u^{ - 2}}du} \cr & {\text{integrate by using }}\int_a^b {{u^n}} du = \left( {\frac{{{u^{n + 1}}}}{{n + 1}}} \right)_a^b \cr & = \frac{1}{2}\left( {\frac{{{u^{ - 1}}}}{{ - 1}}} \right)_4^{3 + {e^2}} \cr & = - \frac{1}{2}\left( {\frac{1}{u}} \right)_4^{3 + {e^2}} \cr & or{\text{ by the properties of the integrals}} \cr & = \frac{1}{2}\left( {\frac{1}{u}} \right)_{3 + {e^2}}^4 \cr & {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr & = \frac{1}{2}\left( {\frac{1}{4} - \frac{1}{{3 + {e^2}}}} \right) \cr & {\text{multiplying}} \cr & = \frac{1}{8} - \frac{1}{{2\left( {3 + {e^2}} \right)}} \cr}