Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.5 The Method of Partial Fractions - Exercises - Page 423: 34


$$\frac{1}{2} \ln |x+1|+\frac{1}{4} \ln \left|x^{2}+1\right|-\frac{1}{2} \tan ^{-1} x+C$$

Work Step by Step

Given $$\int \frac{x^{2}}{(x+1)\left(x^{2}+1\right)} d x$$ Since \begin{align*} \frac{x^2}{(x+1)\left(x^{2}+1\right)}&=\frac{A}{x+1}+\frac{Bx+C}{x^{2}+1}\\ &=\frac{A(x^{2}+1)+B(x-1)}{(x+1)\left(x^{2}+1\right)}\\ x^2&=A(x^{2}+1)+B(x+1) \end{align*} \begin{align*} \text{at } x&=-1 \ \ \ \ \to A=\frac{1}{2} \\ \text{at } x&= 0\ \ \ \ \to C=\frac{-1}{2}\\ \text{at } x&= 1\ \ \ \ \to B =\frac{1}{2} \end{align*} Hence \begin{aligned} \int \frac{x^{2}}{(x+1)\left(x^{2}+1\right)} d x &=\int \frac{\frac{1}{2}}{x+1} d x+\int \frac{\frac{1}{2} x-\frac{1}{2}}{x^{2}+1} d x \\ &=\int \frac{\frac{1}{2}}{x+1} d x+\frac{1}{2} \int \frac{x}{x^{2}+1} d x-\frac{1}{2} \int \frac{1}{x^{2}+1} d x \\ &=\frac{1}{2} \ln |x+1|+\frac{1}{4} \ln \left|x^{2}+1\right|-\frac{1}{2} \tan ^{-1} x+C \end{aligned}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.