To evaluate the integral $\int \sqrt{3x + 4} \, dx$ by substitution, we can follow these steps:

1. Choose a substitution: Let: $u = 3x + 4$ Then, differentiate with respect to $x$: $\frac{du}{dx} = 3 \Rightarrow dx = \frac{du}{3}$

2. Rewrite the integral in terms of $u$: Substitute $u = 3x + 4$ and $dx = \frac{du}{3}$: $\int \sqrt{3x + 4} \, dx = \int \sqrt{u} \cdot \frac{du}{3}$ This simplifies to: $\frac{1}{3} \int u^{1/2} \, du$

3. Integrate: Now, apply the power rule for integration: $\frac{1}{3} \int u^{1/2} \, du = \frac{1}{3} \cdot \frac{u^{3/2}}{3/2} = \frac{1}{3} \cdot \frac{2}{3} u^{3/2} = \frac{2}{9} u^{3/2}$

4. Substitute back $u = 3x + 4$: Substitute $u = 3x + 4$ back into the expression: $\frac{2}{9} (3x + 4)^{3/2} + C$ where $C$ is the constant of integration.

Final Answer:

$\int \sqrt{3x + 4} \, dx = \frac{2}{9} (3x + 4)^{3/2} + C$

Would you like further details or have any questions?

Here are some related questions for further exploration:

1. How would the approach change if the integrand were $\sqrt{3x - 4}$ instead?
2. Can you find the integral $\int (3x + 4)^{n} \, dx$ for a general exponent $n$?
3. How does this substitution method differ from using trigonometric substitutions?
4. What would be the result if the integral was $\int \frac{1}{\sqrt{3x + 4}} \, dx$ instead?
5. Can you verify the answer by differentiating $\frac{2}{9} (3x + 4)^{3/2} + C$?

Tip: When choosing a substitution, try to simplify the integrand to a form where you can apply basic integration rules easily.