Calculus 8th Edition

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

Chapter 4 - Integrals - 4.5 The Substitution Rule - 4.5 Exercises - Page 346: 3


$$\int x^2\sqrt{x^3+1}dx=\frac{2}{9}(x^3+1)^{3/2}$$

Work Step by Step

To find the integral $\int x^2\sqrt{x^3+1}dx$ we will use substitution $u=x^3+1$ which gives us $du=3x^2dx\Rightarrow \frac{1}{3}du=x^2dx.$ Putting this into our integral we get: $$\int x^2\sqrt{x^3+1}dx=\int\sqrt u\cdot\frac{1}{3}du=\frac{1}{3}\int u^{1/2}du=\frac{1}{3}\frac{u^{3/2}}{\frac{3}{2}}=\frac{1}{3}\frac{2}{3}u^{3/2}=\frac{2}{9}u^{3/2}$$ Now we have to express solution in terms of $x$ by expressing $u$ in terms of $x$: $$\int x^2\sqrt{x^3+1}dx=\frac{2}{9}u^{3/2}=\frac{2}{9}(x^3+1)^{3/2}$$
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