Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.1 - Using Basic Integration Formulas - Exercises 8.1 - Page 448: 26


$$\int \frac{6 d y}{\sqrt{y}(1+y)}=12 \tan ^{-1}(\sqrt{y})+c $$

Work Step by Step

Given $$\int \frac{6 d y}{\sqrt{y}(1+y)}$$ Let $$ y=t^2 \Rightarrow dy=2t dt $$ So, we get \begin{aligned} I&=6 \int \frac{2 t d t}{t\left(1+t^{2}\right)}\\ &=12 \quad \int \frac{d t}{1+t^{2}}\\ &=12 \tan^{-1}t\\ &=12 \tan ^{-1}(\sqrt{y})+c\end{aligned}
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