Consider the integral \( I = \int x \sqrt{1 + 2x} \, dx \).
Let \( u = 1 + 2x \). This implies \( du = 2 \, dx \) and \( x = \frac{u - 1}{2} \).
Substituting these into the integral yields \( I = \int \frac{u - 1}{2} \sqrt{u} \cdot \frac{1}{2} \, du = \frac{1}{4} \int (u - 1) u^{\frac{1}{2}} \, du \).
Simplifying the integrand gives \( I = \frac{1}{4} \int \left(u^{\frac{3}{2}} - u^{\frac{1}{2}}\right) \, du \).
Separating the terms, we get \( I = \frac{1}{4} \left( \int u^{\frac{3}{2}} \, du - \int u^{\frac{1}{2}} \, du \right) \).
Integrating each term individually, we have \( \int u^{\frac{3}{2}} \, du = \frac{2}{5} u^{\frac{5}{2}} \) and \( \int u^{\frac{1}{2}} \, du = \frac{2}{3} u^{\frac{3}{2}} \). Substituting these back into the expression for \( I \): \( I = \frac{1}{4} \left(\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}}\right) \).
Finally, simplifying and substituting \( u = 1 + 2x \) back results in \( I = \frac{1}{10} (1 + 2x)^{\frac{5}{2}} - \frac{1}{6} (1 + 2x)^{\frac{3}{2}} + C \).
The solution is: \[ \int x \sqrt{1 + 2x} \, dx = \frac{1}{10} (1 + 2x)^{\frac{5}{2}} - \frac{1}{6} (1 + 2x)^{\frac{3}{2}} + C. \]