## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 5: Integrals - Practice Exercises - Page 308: 51

#### 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}

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