Question

Find: \[ \int x \sqrt{1 + 2x} \, dx \]

Hint:

N/A

Answer 1

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: \[ I = \frac{1}{4} \int \left(u^{\frac{3}{2}} - u^{\frac{1}{2}}\right) \, du. \] 

Split the integral: \[ I = \frac{1}{4} \left( \int u^{\frac{3}{2}} \, du - \int u^{\frac{1}{2}} \, du \right). \] 

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 \( u = 1 + 2x \): \[ I = \frac{1}{10} (1 + 2x)^{\frac{5}{2}} - \frac{1}{6} (1 + 2x)^{\frac{3}{2}} + C. \] 
Answer: \[ \int x \sqrt{1 + 2x} \, dx = \frac{1}{10} (1 + 2x)^{\frac{5}{2}} - \frac{1}{6} (1 + 2x)^{\frac{3}{2}} + C. \]