Answer
$x-\dfrac{x^3}{3}-\dfrac{x^6}{2}+C$
Work Step by Step
Since, we have $\int (1-x^2-3x^5) dx=x-\dfrac{x^3}{3}-\dfrac{x^6}{2}+C$
and
$\dfrac{d}{dx}(x-\dfrac{x^3}{3}-\dfrac{x^6}{2}+C)=1-x^2-3x^5$
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 4 - Section 4.8 - Antiderivatives - Exercises - Page 272: 30
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