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

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myininayaBest ResponseYou've already chosen the best response.1
\[g(t)=\frac{5+t+t^2}{t^\frac{1}{2}}=\frac{5}{t^\frac{1}{2}}+\frac{t}{t^\frac{1}{2}}+\frac{t^2}{t^\frac{1}{2}}=5t^{\frac{1}{2}}+t^{1\frac{1}{2}}+t^{2\frac{1}{2}}\] \[g(t)=5t^{\frac{1}{2}}+t^\frac{1}{2}+t^{\frac{3}{2}}\] Find the antiderivative of each term using \[f(t)=t^n => F(t)=\frac{t^{n+1}}{n+1} \text{ where } F'=f \text{ and } n \neq 1\]
 one year ago
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