Answer
$$\frac{1}{6}$$
Work Step by Step
$$\eqalign{
& \int_0^{\ln 5} {{e^r}{{\left( {3{e^r} + 1} \right)}^{ - 3/2}}} dr \cr
& {\text{use substitution}}{\text{: }} \cr
& {\text{ }}u = 3{e^r} + 1,{\text{ so that }}\frac{{du}}{{dr}} = 3{e^r},\,\,\,dr = \frac{{du}}{{3{e^r}}} \cr
& {\text{the new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}r = \ln 5,{\text{ }}u = 3{e^{\ln 5}} + 1 = 16 \cr
& \,\,\,\,\,\,{\text{If }}r = 0,{\text{ }}u = 3{e^0} + 1 = 4 \cr
& {\text{Then}} \cr
& \int_0^{\ln 5} {{e^r}{{\left( {3{e^r} + 1} \right)}^{ - 3/2}}} dr = \int_4^{16} {{e^r}{u^{ - 3/2}}} \left( {\frac{{du}}{{3{e^r}}}} \right) \cr
& = \frac{1}{3}\int_4^{16} {{u^{ - 3/2}}} du \cr
& {\text{integrating}} \cr
& = \frac{1}{3}\left( {\frac{{{u^{ - 1/2}}}}{{ - 1/2}}} \right)_4^{16} \cr
& = - \frac{2}{3}\left( {\frac{1}{{\sqrt u }}} \right)_4^{16} \cr
& {\text{use the fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = - \frac{2}{3}\left( {\frac{1}{{\sqrt {16} }} - \frac{1}{{\sqrt 4 }}} \right) \cr
& {\text{simplifying, we get:}} \cr
& = - \frac{2}{3}\left( {\frac{1}{4} - \frac{1}{2}} \right) \cr
& = \frac{1}{6} \cr} $$
