To solve the integral

$\int_1^2 \frac{1}{(3 - 5x)^2} \, dx$

we can proceed with a substitution method. Here are the steps:

Step 1: Set up a substitution

Let $u = 3 - 5x$ Then, $\frac{du}{dx} = -5 \Rightarrow dx = -\frac{1}{5} \, du.$

Step 2: Adjust the limits of integration

When $x = 1$: $u = 3 - 5 \cdot 1 = -2.$ When $x = 2$: $u = 3 - 5 \cdot 2 = -7.$

So the integral becomes $\int_{-2}^{-7} \frac{1}{u^2} \cdot \left(-\frac{1}{5}\right) \, du.$

This simplifies to $-\frac{1}{5} \int_{-2}^{-7} \frac{1}{u^2} \, du.$

Step 3: Simplify the integral

Rewrite the integral in terms of $u$: $-\frac{1}{5} \int_{-2}^{-7} u^{-2} \, du.$

Step 4: Integrate $u^{-2}$

The antiderivative of $u^{-2}$ is $\int u^{-2} \, du = -\frac{1}{u}.$

Thus, $-\frac{1}{5} \int_{-2}^{-7} u^{-2} \, du = -\frac{1}{5} \left[-\frac{1}{u}\right]_{-2}^{-7}.$

Step 5: Evaluate the definite integral

Now substitute the limits: $-\frac{1}{5} \left[-\frac{1}{-7} + \frac{1}{-2}\right] = -\frac{1}{5} \left(\frac{1}{7} - \frac{1}{2}\right).$

Step 6: Simplify the result

Combine the terms inside the parentheses: $-\frac{1}{5} \left(\frac{1}{7} - \frac{1}{2}\right) = -\frac{1}{5} \left(\frac{2 - 7}{14}\right) = -\frac{1}{5} \cdot \frac{-5}{14} = \frac{1}{14}.$

Final Answer

$\int_1^2 \frac{1}{(3 - 5x)^2} \, dx = \frac{1}{14}.$

Would you like more details or have any questions on any step?

Here are 5 related questions for practice:

1. How would the answer change if the limits of integration were from 0 to 1 instead?
2. What if the integrand was $\frac{1}{(3 + 5x)^2}$ instead of $\frac{1}{(3 - 5x)^2}$?
3. How can we solve the integral $\int \frac{1}{(ax + b)^2} \, dx$ for general constants $a$ and $b$?
4. What happens if the power in the denominator is $3$ instead of $2$, i.e., $\frac{1}{(3 - 5x)^3}$?
5. How would we approach the integral $\int_1^2 \frac{1}{3 - 5x} \, dx$ without the square?

Tip: Always double-check substitution steps, especially when changing limits of integration.