The integral provided is:

$\int \frac{1}{(3x - 4)^2} \, dx$

Solution:

1. Let $u = 3x - 4$, so that $du = 3 \, dx$, or equivalently $dx = \frac{du}{3}$.

2. Substituting $u$ into the integral: $\int \frac{1}{(3x - 4)^2} \, dx = \int \frac{1}{u^2} \cdot \frac{du}{3}$

3. Simplify the constants: $\frac{1}{3} \int \frac{1}{u^2} \, du$

4. The integral of $\frac{1}{u^2}$ is $-\frac{1}{u}$: $\frac{1}{3} \cdot \left( -\frac{1}{u} \right) + C = -\frac{1}{3u} + C$

5. Substitute back $u = 3x - 4$: $-\frac{1}{3(3x - 4)} + C$

Final Answer:

$\int \frac{1}{(3x - 4)^2} \, dx = -\frac{1}{3(3x - 4)} + C$

Would you like further details or a breakdown of any step?

Related Questions:

1. How can substitution simplify integrals like this one?
2. What is the general formula for integrating powers of $(ax + b)^n$?
3. How would this integral change if it were $(3x - 4)^3$ in the denominator?
4. What is the antiderivative of $\frac{1}{(ax + b)^2}$?
5. How do indefinite and definite integrals differ in approach?

Tip:

When dealing with rational functions, substitution is often the most efficient method to simplify and solve the integral!