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: 30


$$\int x^3\sqrt{x^2+1}dx=\frac{(x^2+1)^{3/2}(3x^2-2)}{15}+c$$

Work Step by Step

To solve the integral $\int x^3\sqrt{x^2+1}dx$ we will use substitution $x^2+1=t$ which gives us $x^2=t-1$and $2xdx=dt\Rightarrow xdx=\frac{dt}{2}$. Putting this into the integral we get: $$\int x^3\sqrt{x^2+1}dx=\int \sqrt{x^2+1}\cdot x^2\cdot xdx= \int\sqrt t(t-1)\frac{dt}{2}=\frac{1}{2}\int t^{3/2}dt-\frac{1}{2}\int t^{1/2}dt= \frac{1}{2}\frac{t^{5/2}}{\frac{5}{2}}-\frac{1}{2}\frac{t^{3/2}}{\frac{3}{2}}+c=\frac{1}{2}\frac{2}{5}t^{5/2}-\frac{1}{2}\frac{3}{2}t^{3/2}+c= \frac{1}{5}t^{5/2}-\frac{1}{3}t^{3/2}+c= \frac{t^{3/2}(3t-5)}{15}+c$$ where $c$ is arbitrary constant. Now we have to express solution in terms of $x$: $$\int x^3\sqrt{x^2+1}dx=\frac{t^{3/2}(3t-5)}{15}+c=\frac{(x^2+1)^{3/2}(3(x^2+1)-5)}{15}+c= \frac{(x^2+1)^{3/2}(3x^2-2)}{15}+c$$
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