Answer
$$G(t) =2t^{1/2}+\frac{2}{3}t^{3/2}+\frac{2}{5}t^{5/2} +C$$
Work Step by Step
Given $$g(t) =\frac{1+t+t^2}{\sqrt{t}}= t^{-1/2}+t^{1/2}+t^{3/2}$$
Then by using if $f(x)=x^{\alpha}\ \to\ \ F(x) = \frac{x^{\alpha+1}}{\alpha+1}+C$
\begin{align*}
G(t) &= \frac{1}{1/2} t^{1/2}+\frac{1}{3/2}t^{3/2}+\frac{1}{5/2}t^{5/2}+C\\
&=2t^{1/2}+\frac{2}{3}t^{3/2}+\frac{2}{5}t^{5/2} +C
\end{align*}
To check
\begin{align*}
G'(t) &=t^{-1/2}+t^{1/2}+t^{3/2} +C \\
&=\frac{1+t+t^2}{\sqrt{t}}\\
&=g(t)
\end{align*}
