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 $7.07 - 4 = 3.07$.

Continuing from where you left off, which is correct so far...
\begin{align} 2x^{3/2}\vert_1^2 + 2x^{1/2}\vert_1^2&=[2(2\sqrt{2})-2(1)]+[2(\sqrt{2})-2(1)] \\
&=4\sqrt{2}-2+2\sqrt{2}-2 \\
&=6\sqrt{2}-4\end{align}
