Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 10 - Introduction to Differential Equations - 10.1 Solving Differential Equations - Exercises - Page 505: 25


$$x=\tan \left(\frac{t^{2}}{2}+t+C\right) $$

Work Step by Step

Since \begin{aligned} \frac{d x}{d t} &=(t+1)\left(x^{2}+1\right) \\ \frac{d x}{x^{2}+1} &=\frac{(t+1)\left(x^{2}+1\right)}{x^{2}+1} \\ \frac{1}{x^{2}+1} d x &=(t+1) d t \end{aligned} Then \begin{aligned} \int \frac{1}{x^{2}+1} d x &=\int(t+1) d t \\ \tan ^{-1} x &=\frac{t^{2}}{2}+t+C \\ x &=\tan \left(\frac{t^{2}}{2}+t+C\right) \end{aligned}
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