Answer
$$\frac{4}{5}$$
Work Step by Step
$$\eqalign{
& \int_4^6 {\frac{2}{{{{\left( {2x - 7} \right)}^2}}}} dx \cr
& {\text{use substitution}}{\text{. Let }}u = 2x - 7,{\text{ so that }}du = 2dx \cr
& {\text{the new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = 6,{\text{ then }}u = 2\left( 6 \right) - 7 = 5 \cr
& \,\,\,\,\,\,{\text{If }}x = 4,{\text{ then }}u = 2\left( 4 \right) - 7 = 1 \cr
& {\text{Then}} \cr
& \int_4^6 {\frac{2}{{{{\left( {2x - 7} \right)}^2}}}} dx = \int_1^5 {\frac{{du}}{{{u^2}}}} \cr
& {\text{use }}\frac{1}{{{u^n}}} = {u^{ - n}} \cr
& = \int_1^5 {{u^{ - 2}}} du \cr
& {\text{use }}\int_a^b {{u^n}} du = \left( {\frac{{{u^{n + 1}}}}{{n + 1}}} \right)_a^b \cr
& = \left( {\frac{{{u^{ - 2 + 1}}}}{{ - 2 + 1}}} \right)_1^5 \cr
& = \left( {\frac{{{u^{ - 1}}}}{{ - 1}}} \right)_1^5 \cr
& = \left( { - \frac{1}{u}} \right)_1^5 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = - \frac{1}{5} + \frac{1}{1} \cr
& = \frac{4}{5} \cr} $$
