Answer
$\dfrac{-1}{x}-\dfrac{1}{2x^2}+C$
Work Step by Step
As we know $\int x^n dx=\dfrac{x^{n+1}}{n+1}+C$
Thus, we have $\int x^{-3}(x+1) dx=\int (x^{-2}+x^{-3})dx$
or, $=-x^{-1}-\dfrac{x^{-2}}{2}+C$
or, $=\dfrac{-1}{x}-\dfrac{1}{2x^2}+C$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 4: Applications of Derivatives - Section 4.7 - Antiderivatives - Exercises 4.7 - Page 239: 32
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