University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Practice Exercises - Page 474: 15


$4 \ln |x|-\dfrac{1}{2} \ln (x^2+1)+4 \tan^{-1} x+c$

Work Step by Step

Integration by parts formula suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$ Solve $\int \dfrac{3x^2+4x+4}{x^3+x}dx$ Write the integral into partial fractions as follows: $\int \dfrac{3x^2+4x+4}{x^3+x}dx=\int \dfrac{4}{x} dx+\int \dfrac{-(x+4)}{(x^2+1)}dx$ Plug $x^2+1=a \implies da=(2x) dx$ This implies that $4 \ln |x|-\dfrac{1}{2}\int \dfrac{da}{a}+(4) \int \dfrac{dx}{(x^2+1)}=4 \ln |x|-\dfrac{1}{2} \ln (x^2+1)+4 \tan^{-1} x+c$
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