Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.4 Integration of Rational Functions by Partial Fractions - 7.4 Exercises - Page 541: 44


$$\int \frac{d x}{(1+\sqrt{x})^{2}}=2\left[\ln (1+\sqrt{x})+\frac{1}{1+\sqrt{x}}\right]+C$$

Work Step by Step

Given $$\int \frac{d x}{(1+\sqrt{x})^{2}}$$ So, \begin{array}{l}{u=1+\sqrt{x}, u-1=\sqrt{x}} \\ {u=1+x^{1 / 2}} \\ {d u=\frac{1}{2} x^{-1 / 2} d x} \\ {d u=\frac{1}{2 \sqrt{x}} d x} \\ {d x=2 \sqrt{x} d u} \\ {d x=2(u-1) d u}\end{array} So, we get \begin{array}{l}{I=\int \frac{d x}{(1+\sqrt{x})^{2}}}\\{=2 \int \frac{u-1}{u^{2}} d u} \\ {=2 \int \frac{u}{u^{2}}-\frac{1}{u^{2}} d u} \\ {=2 \int \frac{1}{u}-\frac{1}{u^{2}} d u} \\ {=2\left(\ln u+\frac{1}{u}\right)+C} \\ {=2\left[\ln (1+\sqrt{x})+\frac{1}{1+\sqrt{x}}\right]+C}\end{array}
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