Answer
$-e^{-t^2} +c$
Work Step by Step
Solve $\int 2te^{-t^2}dt$
Let us $p=-t^2$ and $dp=-2t dt$
This implies $\int 2te^{-t^2}dt=-\int e^p dp $
$=-e^{p} +c$
$=-e^{-t^2} +c$
Hence,$\int 2te^{-t^2}dt=-e^{-t^2} +c$
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 401: 17
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