Answer
$$\ln \left( {\frac{9}{{25}}} \right)$$
Work Step by Step
$$\eqalign{
& \int_0^4 {\frac{{2t}}{{{t^2} - 25}}} dt \cr
& {\text{Use substitution:}}\cr
&{\text{Let }}u = {t^2} - 25,{\text{ so that }}du = 2tdt \cr
& {\text{The new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}t = 4,{\text{ then }}u = {\left( 4 \right)^2} - 25 = - 9 \cr
& \,\,\,\,\,\,{\text{If }}t = 0,{\text{ then }}u = {\left( 4 \right)^2} - 25 = - 25 \cr
& {\text{Then}} \cr
& \int_0^4 {\frac{{2t}}{{{t^2} - 5}}} dt = \int_{ - 25}^{ - 9} {\frac{{du}}{u}} \cr
& {\text{integrate}} \cr
& = \left( {\ln \left| u \right|} \right)_{ - 25}^{ - 9} \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = \ln \left| { - 9} \right| - \ln \left| { - 25} \right| \cr
& {\text{simplifying, we get:}} \cr
& = \ln \left( {\frac{9}{{25}}} \right) \cr} $$
