Let \( I = \int x \sqrt{1 + 2x} \, dx \).
Using substitution, let \( u = 1 + 2x \). Then,
\[
du = 2 \, dx \quad \text{and} \quad x = \frac{u - 1}{2}.
\]
Substitute into the integral:
\[
I = \int \frac{u - 1}{2} \sqrt{u} \cdot \frac{1}{2} \, du = \frac{1}{4} \int (u - 1) u^{\frac{1}{2}} \, du.
\]
Simplify:
\[
I = \frac{1}{4} \left( \int u^{\frac{3}{2}} \, du - \int u^{\frac{1}{2}} \, du \right).
\]
Integrate each term:
\[
\int u^{\frac{3}{2}} \, du = \frac{2}{5} u^{\frac{5}{2}}, \quad \int u^{\frac{1}{2}} \, du = \frac{2}{3} u^{\frac{3}{2}}.
\]
Substitute back:
\[
I = \frac{1}{4} \left(\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}}\right).
\]
Simplify and substitute \( u = 1 + 2x \):
\[
I = \frac{1}{10} (1 + 2x)^{\frac{5}{2}} - \frac{1}{6} (1 + 2x)^{\frac{3}{2}} + C.
\]
Final Answer:
\[
\boxed{\int x \sqrt{1 + 2x} \, dx = \frac{1}{10} (1 + 2x)^{\frac{5}{2}} - \frac{1}{6} (1 + 2x)^{\frac{3}{2}} + C.}
\]