## lysa_nicole4794 3 years ago Find the most general antiderivative of the function. g(t) = (5 + t + t^2)/ square root of t

1. myininaya

$g(t)=\frac{5+t+t^2}{t^\frac{1}{2}}=\frac{5}{t^\frac{1}{2}}+\frac{t}{t^\frac{1}{2}}+\frac{t^2}{t^\frac{1}{2}}=5t^{-\frac{1}{2}}+t^{1-\frac{1}{2}}+t^{2-\frac{1}{2}}$ $g(t)=5t^{-\frac{1}{2}}+t^\frac{1}{2}+t^{\frac{3}{2}}$ 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

thanks!