Answer
$$8$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{{36dx}}{{{{\left( {2x + 1} \right)}^3}}}} \cr
& {\text{use the substitution method }} \cr
& u = 2x + 1,\,\,\,\,du = 2dx,\,\,\,\,\,dx = \frac{{du}}{2} \cr
& {\text{Then}}{\text{,}} \cr
& \int {\frac{{36}}{{{{\left( {2x + 1} \right)}^3}}}} dx = \int {\frac{{36}}{{{u^3}}}} \left( {\frac{{du}}{2}} \right) \cr
& = 18\int {{u^{ - 3}}} du \cr
& {\text{integrate}} \cr
& = 18\left( {\frac{{{u^{ - 2}}}}{{ - 2}}} \right) + C \cr
& = - 9{u^{ - 2}} + C \cr
& {\text{write in terms of }}u;{\text{ replace }}t = 2x + 1 \cr
& = - \frac{9}{{{{\left( {2x + 1} \right)}^2}}} + C \cr
& {\text{Then}}{\text{,}} \cr
& \int_0^1 {\frac{{36dx}}{{{{\left( {2x + 1} \right)}^3}}}} = \left( { - \frac{9}{{{{\left( {2x + 1} \right)}^2}}}} \right)_0^1 \cr
& {\text{Evaluating, we get:}} \cr
& = - \frac{9}{{{{\left( {2\left( 1 \right) + 1} \right)}^2}}} + \frac{9}{{{{\left( {2\left( 0 \right) + 1} \right)}^2}}} \cr
& = - \frac{9}{9} + \frac{9}{1} \cr
& = 8 \cr} $$
