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: 5

Answer

$$\int\frac{x^3}{(x^4-5)^2}dx=-\frac{1}{4(x^4-5)}+c$$

Work Step by Step

To solve the integral $$\int\frac{x^3}{(x^4-5)^2}dx$$ we will use subtitution $u=x^4-5$ which gives us $du=4x^3dx\Rightarrow \frac{1}{4}du=x^3dx.$ Putting this into the integral we get: $$\int\frac{x^3}{(x^4-5)^2}dx=\int \frac{1}{u^2}\frac{1}{4}du=\frac{1}{4}\int u^{-2}du=\frac{1}{4}(\frac{u^{-1}}{-1}+c_{1})=-\frac{1}{4}\frac{1}{u}+c$$ where $c$ is arbitrary constant. Now we have to express out solution in terms of $x$ by expressing $u$ in terms of $x$: $$\int\frac{x^3}{(x^4-5)^2}dx=-\frac{1}{4u}+c=-\frac{1}{4(x^4-5)}+c$$
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