To integrate the expression $\frac{1}{(x - 11)(x - 21)}$, we can use partial fraction decomposition.

Step 1: Partial Fraction Decomposition

We express the fraction as a sum of two simpler fractions: $\frac{1}{(x - 11)(x - 21)} = \frac{A}{x - 11} + \frac{B}{x - 21}$ Now, multiply both sides by $(x - 11)(x - 21)$ to eliminate the denominators: $1 = A(x - 21) + B(x - 11)$ Now, expand both terms: $1 = A(x) - 21A + B(x) - 11B$ $1 = (A + B)x - (21A + 11B)$

Step 2: Solve for $A$ and $B$

For the equation $1 = (A + B)x - (21A + 11B)$ to hold, the coefficient of $x$ must be 0, and the constant term must be 1. This gives the system of equations: $A + B = 0$ $-21A - 11B = 1$

From $A + B = 0$, we have $B = -A$. Substitute this into the second equation: $-21A - 11(-A) = 1$ $-21A + 11A = 1$ $-10A = 1$ $A = -\frac{1}{10}$ Since $B = -A$, we get: $B = \frac{1}{10}$

Step 3: Rewrite the Integral

Now that we know $A = -\frac{1}{10}$ and $B = \frac{1}{10}$, we rewrite the integrand: $\frac{1}{(x - 11)(x - 21)} = \frac{-\frac{1}{10}}{x - 11} + \frac{\frac{1}{10}}{x - 21}$

Step 4: Integrate

Now, integrate each term: $\int \frac{1}{(x - 11)(x - 21)} dx = -\frac{1}{10} \int \frac{1}{x - 11} dx + \frac{1}{10} \int \frac{1}{x - 21} dx$ The integral of $\frac{1}{x - a}$ is $\ln|x - a|$, so we get: $= -\frac{1}{10} \ln|x - 11| + \frac{1}{10} \ln|x - 21| + C$

Final Answer:

$\int \frac{1}{(x - 11)(x - 21)} dx = \frac{1}{10} \left( \ln|x - 21| - \ln|x - 11| \right) + C$ You can also combine the logarithms: $= \frac{1}{10} \ln \left| \frac{x - 21}{x - 11} \right| + C$

Let me know if you'd like further clarification or more details! Here are 5 related questions to explore further:

1. How would this approach change if we had more terms in the denominator, like $(x-11)(x-21)(x-31)$?
2. What are some techniques to solve integrals involving higher powers of binomials, such as $(x-11)^2$?
3. How does the constant $C$ impact the interpretation of the integral in real-world applications?
4. Can we solve this type of integral without using partial fractions? If so, how?
5. How do partial fractions help in solving integrals of rational functions in general?

Tip: When you encounter integrals with rational functions, always try to use partial fraction decomposition as it can simplify the process!