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 347: 75


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

Work Step by Step

We will first separate this integral into two new ones by additivity of integrals: $$\int\frac{1+x}{1+x^2}dx=\int\frac{1}{1+x^2}dx+\int\frac{x}{1+x^2}dx$$ First integral: $$\int\frac{1}{1+x^2}dx=\arctan x+c$$ To evaluate the second integral we will use substitution $1+x^2=t$ which gives us $2xdx=dt\Rightarrow xdx=\frac{dt}{2}$. Putting this into the integral we get: $$\int\frac{x}{1+x^2}dx=\int\frac{1}{t}\frac{dt}{2}=\frac{1}{2}\ln|t|+c$$ Now we have to express solution in terms of $x$: $$\int\frac{x}{1+x^2}dx=\frac{1}{2}\ln|t|+c=\frac{1}{2}\ln(1+x^2)+c$$ Finally we have: $$\int{1+x}{1+x^2}dx=\arctan x+\frac{1}{2}\ln(1+x^2)+c$$
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