To expand \(\frac{1 + 3x}{\sqrt{1 + 2x}}\), we first consider the expansion of \((1 + 2x)^{-\frac{1}{2}}\).

Using the binomial expansion for \((1 + u)^n\) where \(n = -\frac{1}{2}\) and \(u = 2x\), we have:

\((1 + 2x)^{-\frac{1}{2}} = 1 - x + \frac{3}{2}x^2 + \cdots\)

Now, multiply this expansion by \(1 + 3x\):

\((1 + 3x)(1 - x + \frac{3}{2}x^2) = 1 + 3x - x - 3x^2 + \frac{3}{2}x^2 + \cdots\)

Combine like terms:

\(1 + 2x - \frac{3}{2}x^2\)

Thus, the expansion up to the term in \(x^2\) is \(1 + 2x - \frac{3}{2}x^2\).