Answer
$$4{\tan ^{ - 1}}\left( {\frac{\pi }{4}} \right)$$
Work Step by Step
$$\eqalign{
& \int_1^{{e^{\pi /4}}} {\frac{{4dt}}{{t\left( {1 + {{\ln }^2}t} \right)}}} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = \ln t,{\text{ so that }}du = \frac{1}{t}dt \cr
& {\text{the new limits on }}t{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}t = {e^{\pi /4}},{\text{ then }}u = \ln {e^{\pi /4}} = \pi /4 \cr
& \,\,\,\,\,\,{\text{If }}t = 1,{\text{ then }}u = \ln 1 = 0 \cr
& {\text{then}} \cr
& \int_1^{{e^{\pi /4}}} {\frac{{4dt}}{{t\left( {1 + {{\ln }^2}t} \right)}}} = 4\int_0^{\pi /4} {\frac{{du}}{{1 + {u^2}}}} \cr
& {\text{integrate by using the formula }}\int {\frac{{du}}{{{a^2} + {u^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 1 \cr
& = 4\left( {{{\tan }^{ - 1}}\left( u \right)} \right)_0^{\pi /4} + C \cr
& = 4{\tan ^{ - 1}}\left( {\frac{\pi }{4}} \right) - 4{\tan ^{ - 1}}\left( 0 \right) \cr
& = 4{\tan ^{ - 1}}\left( {\frac{\pi }{4}} \right) \cr} $$
