Make the following substitution:
$\displaystyle u = 4+ 3x$
$\displaystyle du = 3dx \: \: \Rightarrow \: \: dx = \frac{du}{3}$
Also, we must change our limits of our integral:
$\displaystyle u(8) = 4 + 3(8) = 28$
$\displaystyle u(2) = 4 + 3(2) = 10$
So substituting all these in:
$\displaystyle = \int_{u(2)}^{u(8)}\! \sqrt{u} \frac{du}{3} \: = \: \frac{1}{3} \int_{10}^{28} u^{\frac{1}{2}}du$
From where we last left off:
$\displaystyle \frac{1}{3} \int_{10}^{28} u^{\frac{1}{2}}du = \frac{1}{3} \cdot \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \bigg|_{10}^{28} \: \: = \: \: \frac{2}{9} u^{\frac{3}{2}} \bigg|_{10}^{28} \: \: = \: \: \frac{2}{9} \left( 28^{\frac{3}{2}} - 10^{\frac{3}{2}} \right) \: \approx \: 25.898$
Just as you had