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: 74

Answer

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

Work Step by Step

To evaluate the integral $$\int\frac{x}{x^2+4}dx$$ we will use substitution $x^2+4=t$ which gives us $2xdx=dt\Rightarrow xdx=\frac{dt}{2}$. Putting this into the integral we get: $$\int\frac{x}{x^2+4}dx=\int\frac{1}{t}\frac{dt}{2}=\frac{1}{2}\int\frac{1}{t}dt=\frac{1}{2}\ln|t|+c$$ where $c$ is arbitrary constant. Now we have to express solution in terms of $x$: $$\int\frac{x}{x^2+4}dx=\frac{1}{2}\ln|t|+c=\frac{1}{2}\ln(x^2+4)+c$$
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