Answer
$F\left( t \right) = \frac{4}{3}t\sqrt t - 8\sqrt t + 3t + C$
Work Step by Step
$$\eqalign{
& f\left( t \right) = \frac{{2t - 4 + 3\sqrt t }}{{\sqrt t }} \cr
& {\text{First use the radical property }}\sqrt m = {m^{1/2}} \cr
& f\left( t \right) = \frac{{2t - 4 + 3{t^{1/2}}}}{{{t^{1/2}}}} \cr
& {\text{Distribute the numerator}} \cr
& f\left( t \right) = \frac{{2t}}{{{t^{1/2}}}} - \frac{4}{{{t^{1/2}}}} + \frac{{3{t^{1/2}}}}{{{t^{1/2}}}} \cr
& f\left( t \right) = 2{t^{1/2}} - 4{t^{ - 1/2}} + 3 \cr
& {\text{Using the formulas in Table 2 }}\left( {{\text{see page 358}}} \right){\text{ }} \cr
& {\text{Function: }}cf\left( x \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr
& {\text{Function: }}{x^n}\left( {n \ne - 1} \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr
& {\text{Function: }}f\left( x \right) + h\left( x \right) \to {\text{Particular antiderivative: }}f\left( x \right) + h\left( x \right) \cr
& {\text{And applying the Theorem 1}},{\text{ we obtain}} \cr
& F\left( t \right) = 2\left( {\frac{{{t^{1/2 + 1}}}}{{1/2 + 1}}} \right) - 4\left( {\frac{{{t^{ - 1/2 + 1}}}}{{ - 1/2 + 1}}} \right) + 3t + C \cr
& F\left( t \right) = 2\left( {\frac{{{t^{3/2}}}}{{3/2}}} \right) - 4\left( {\frac{{{t^{1/2}}}}{{1/2}}} \right) + 3t + C \cr
& F\left( t \right) = \frac{4}{3}{t^{3/2}} - 8{t^{1/2}} + 3t + C \cr
& {\text{Use the radical properties}} \cr
& F\left( t \right) = \frac{4}{3}t\sqrt t - 8\sqrt t + 3t + C \cr} $$
