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$f(t) = \frac{t^2 + 9}{t}$ \newline An antiderivative of $f(t)$ is $\frac{1}{2}t^2 + 9\ln(t) + C$

Name: f(t) = (t^2 + 9)/(t) An antiderivative of f(t) is (1)/(2)t^2 + 9ln(t) + C
Uploaded: 2023-11-08T00:52:27-08:00
Duration: 1 min 24 s
Channel: Supreeta N
Description: f(t) = (t^2 + 9)/(t) An antiderivative of f(t) is (1)/(2)t^2 + 9ln(t) + C

          $f(t) = \frac{t^2 + 9}{t}$ \newline An antiderivative of $f(t)$ is $\frac{1}{2}t^2 + 9\ln(t) + C$

Added by Aaron T.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Supreeta N

11/08/2023

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Step 1: The given function is f(t) = (t^2 + 9)/t. Show more…

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f(t)=(t^(2)+9)/(t) An antiderivative of f(t) is (1)/(2)t^(2)+9ln(t)+C t2+9 f(t) t An antiderivative of f(t) is t2+91n(t)+C x

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$f(t) = \frac{t^2 + 9}{t}$ \newline An antiderivative of $f(t)$ is $\frac{1}{2}t^2 + 9\ln(t) + C$

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