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Evaluate the integral using substitution

Use the method of substitution to evaluate the following integral:

    \[ \int x \sqrt{1 + 3x} \, dx. \]


Let

    \begin{align*}  u &= 1 + 3x & \implies && x = \frac{u-1}{3} \\  du &= 3 \, dx & \implies && dx = \frac{1}{3} \, du. \end{align*}

Then, we have

    \begin{align*}  \int x \sqrt{1+3x} \, dx &= \int \left( \frac{u-1}{3} \sqrt{u} \right) \frac{du}{3} \\  &= \frac{1}{9} \int \left( u^{\frac{3}{2}} - u^{\frac{1}{2}} \right) \, du \\  &= \frac{1}{9} \left( \int u^{\frac{3}{2}} \, du - \int u^{\frac{1}{2}} \, du \right)\\  &= \frac{1}{9} \left( \frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}} \right) + C \\  &= \frac{2}{45} (1+3x)^{\frac{5}{2}} - \frac{2}{27} (1+3x)^{\frac{3}{2}} + C. \end{align*}

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