Integrating $(3x + 1) / x^\frac{1}{2}$ I am trying to integrate the following equation, but my answer is different from the textbook and I cannot see where I am going wrong:
\begin{align} \int_1^2\frac{ 3x + 1}{x^{1/2}}dx
&=  \int_1^2 \frac{3x}{x^{1/2}}dx +  \int_1^2 \frac{1}{x^{1/2}}dx \\
&=  \int_1^2 3x^{1/2}dx +  \int_1^2 x^{-1/2}dx\\
&= 2x^{3/2}\vert_1^2 + 2x^{1/2}\vert_1^2 
\end{align}
The book has an answer $6\sqrt{2} - 4$, but according to what I have just done the answer is roughly $7.07 - 4 = 3.07$.

$$\begin{align} \int_1^2 3x^{1/2} \,dx + \int_1^2 x^{-1/2}\,dx & = \frac{3x^{3/2}}{{\frac 32}} + \frac{x^{1/2}}{1/2}\Big|_1^2\\ \\
& = 2 x^{3/2} + 2x^{1/2}\Big|_1^2 \\ \\ 
& = 4\sqrt 2 + 2\sqrt 2 - (2 + 2)\\ \\ & = 6\sqrt 2 - 4\end{align}$$
