Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises: 48


$$\int x^3\sqrt{x^2+1}=\frac{1}{5}\sqrt{(x^2+1)^5}-\frac{1}{3}\sqrt{(x^2+1)^3}+C$$

Work Step by Step

$$A=\int x^3\sqrt{x^2+1}dx$$ $$A=\int x^2\sqrt{x^2+1}(xdx)$$ Let $u=x^2+1$. Then $du=2xdx$. Therefore, $xdx=\frac{1}{2}du$ And since $u=x^2+1$, we have $x^2=u-1$ Substitute into $A$: $$A=\frac{1}{2}\int(u-1)\sqrt udu$$ $$A=\frac{1}{2}\int(u-1)u^{1/2}du$$ $$A=\frac{1}{2}\int(u^{3/2}-u^{1/2})du$$ $$A=\frac{1}{2}(\frac{u^{5/2}}{\frac{5}{2}}-\frac{u^{3/2}}{\frac{3}{2}})+C$$ $$A=\frac{1}{2}(\frac{2\sqrt{u^5}}{5}-\frac{2\sqrt{u^3}}{3})+C$$ $$A=\frac{\sqrt{u^5}}{5}-\frac{\sqrt{u^3}}{3}+C$$ $$A=\frac{1}{5}\sqrt{(x^2+1)^5}-\frac{1}{3}\sqrt{(x^2+1)^3}+C$$
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