Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 5 - Integration - 5.5 Substitution Rule - 5.5 Exercises: 40

Answer

\[ = \frac{4}{5}\]

Work Step by Step

\[\begin{gathered} \hfill \\ \int_0^2 {\frac{{2x}}{{\,{{\left( {{x^2} + 1} \right)}^2}}}} \,dx \hfill \\ \hfill \\ Let\,\,u = {x^2} + 1\,\,\,{\text{which implies that}}\,\,\,\,\,du = 2xdx \hfill \\ \hfill \\ {\text{Changing limits of integration}} \hfill \\ \hfill \\ x = 2\,\,\,implies\,\,\,\,u = 5 \hfill \\ x = 0\,\,implies\,\,\,\,u = 1 \hfill \\ \hfill \\ apply\,\,the\,\,\,substitution \hfill \\ \hfill \\ \int_1^5 {\frac{{du}}{{{u^2}}}} \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \,\,\left[ { - \frac{1}{u}} \right]_1^5 \hfill \\ \hfill \\ Fundamental\,\,theorem \hfill \\ \hfill \\ = - \frac{1}{5} + \frac{1}{1} = \frac{4}{5} \hfill \\ \end{gathered} \]
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