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lysa_nicole4794

  • 3 years ago

Find the most general antiderivative of the function. g(t) = (5 + t + t^2)/ square root of t

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  1. myininaya
    • 3 years ago
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    \[g(t)=\frac{5+t+t^2}{t^\frac{1}{2}}=\frac{5}{t^\frac{1}{2}}+\frac{t}{t^\frac{1}{2}}+\frac{t^2}{t^\frac{1}{2}}=5t^{-\frac{1}{2}}+t^{1-\frac{1}{2}}+t^{2-\frac{1}{2}}\] \[g(t)=5t^{-\frac{1}{2}}+t^\frac{1}{2}+t^{\frac{3}{2}}\] Find the antiderivative of each term using \[f(t)=t^n => F(t)=\frac{t^{n+1}}{n+1} \text{ where } F'=f \text{ and } n \neq -1\]

  2. lysa_nicole4794
    • 3 years ago
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    thanks!

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