Answer
$$ - \frac{1}{3}$$
Work Step by Step
$$\eqalign{
& \int_1^4 {\frac{{ - 3}}{{{{\left( {2p + 1} \right)}^2}}}} dp \cr
& {\text{use substitution}}{\text{. Let }}u = 2p + 1,{\text{ so that }}du = 2pdp,\,\,\,\,\frac{1}{2}du = dp \cr
& {\text{the new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}p = 4,{\text{ then }}u = 2\left( 4 \right) + 1 = 9 \cr
& \,\,\,\,\,\,{\text{If }}p = 1,{\text{ then }}u = 2\left( 1 \right) + 1 = 3 \cr
& {\text{Then}} \cr
& \int_1^4 {\frac{{ - 3}}{{{{\left( {2p + 1} \right)}^2}}}} dp = \int_3^9 {\frac{{ - 3}}{{{u^2}}}\left( {\frac{1}{2}du} \right)} \cr
& = - \frac{3}{2}\int_3^9 {\frac{1}{{{u^2}}}du} \cr
& {\text{use }}\frac{1}{{{u^n}}} = {u^{ - n}} \cr
& = - \frac{3}{2}\int_3^9 {{u^{ - 2}}du} \cr
& {\text{use }}\int_a^b {{u^n}} du = \left( {\frac{{{u^{n + 1}}}}{{n + 1}}} \right)_a^b \cr
& = - \frac{3}{2}\left( {\frac{{{u^{ - 2 + 1}}}}{{ - 2 + 1}}} \right)_3^9 \cr
& = - \frac{3}{2}\left( {\frac{{{u^{ - 1}}}}{{ - 1}}} \right)_3^9 \cr
& = \frac{3}{2}\left( {\frac{1}{u}} \right)_3^9 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = \frac{3}{2}\left( {\frac{1}{9} - \frac{1}{3}} \right) \cr
& = \frac{3}{2}\left( { - \frac{2}{9}} \right) \cr
& = - \frac{1}{3} \cr} $$
