Answer
$y=x+\ln |x|+2$
Work Step by Step
Here, $y=\int (1+\dfrac{1}{x}) dx$
Now, we have $y=\int (1) dx+\int (\dfrac{1}{x}) dx$
$y=x+\ln |x|+c$
Apply initial conditions $y(1)=3$
we get $c=2$
Thus, we have $y=x+\ln |x|+2$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 7 - Section 7.1 - The Logarithm Defined as an Integral - Exercises - Page 402: 51
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