Write the integral in terms of $u$ and du. Then evaluate.
\[
\int x \sqrt{3 x+4} d x, \quad u=3 x+4
\]
(Use symbolic notation and fractions where needed. Use $C$ for the arbitrary constant. Absorb into $C$ as much as possible.)
\[
\int x \sqrt{3 x+4} d x=
\]

Step 1 ：Let \( u = 3x + 4 \).

Step 2 ：Solve for \( x \) to get \( x = \frac{u}{3} - \frac{4}{3} \).

Step 3 ：Differentiate \( u = 3x + 4 \) with respect to \( x \) to get \( du = 3dx \).

Step 4 ：Solve for \( dx \) to get \( dx = \frac{du}{3} \).

Step 5 ：Substitute \( x \) and \( dx \) in the integral with their expressions in terms of \( u \) and \( du \) to get \( \int \sqrt{u} \left( \frac{u}{3} - \frac{4}{3} \right) du \).

Step 6 ：Evaluate the integral to get \( \frac{2}{45} u^{3/2} - \frac{8}{45} u^{1/2} + \frac{64}{135} u^{1/2} + C \).

Step 7 ：Substitute \( u = 3x + 4 \) back into the integral to get the final answer.

Step 8 ：\(\boxed{\int x \sqrt{3 x+4} dx = \frac{2}{45} (3x + 4)^{3/2} - \frac{8}{45} (3x + 4)^{1/2} + \frac{64}{135} (3x + 4)^{1/2} + C}\)