Integrating Rational Functions with Repeated Roots via Partial Fractions

Concept Overview

This question tests the ability to integrate rational functions where the denominator contains repeated linear factors. The core technique involves decomposing the rational function into simpler fractions using partial fraction decomposition. For repeated roots, each power of the factor, from one up to its multiplicity, requires a separate term in the decomposition, which simplifies the integration process significantly.

Step 1: Set up the partial fraction decomposition. When the denominator of a rational function has a repeated linear factor, say $(ax+b)^n$, the partial fraction decomposition includes terms for each power of this factor from 1 to $n$. For a denominator like $(x-a)^n$, the decomposition will look like:

In our case, let's consider integrating a general rational function $\frac{P(x)}{(x-a)^n Q(x)}$ where $Q(x)$ has distinct roots. The decomposition will be:

where $c_i$ are the distinct roots of $Q(x)$.

Step 2: Clear the denominators and solve for coefficients. To find the coefficients ($A_i$ and $B_i$), we multiply both sides of the partial fraction equation by the original denominator $(x-a)^n Q(x)$. This results in a polynomial identity.

The coefficients can be found by substituting strategic values of $x$ (especially the roots of the denominator) or by comparing coefficients of like powers of $x$. For repeated roots, substituting the root $x=a$ is particularly useful for finding the coefficients of the highest powers.

Step 3: Integrate each term. Once the coefficients are determined, the original integral can be rewritten as a sum of simpler integrals.

The integrals of the terms involving distinct roots are standard logarithmic forms: $\int \frac{B_i}{x-c_i} dx = B_i \ln|x-c_i| + C$.

Step 4: Integrate terms with repeated roots. The terms with repeated roots are integrated using a power rule for integration. For a term $\frac{A_k}{(x-a)^k}$:

Using the power rule $\int u^m du = \frac{u^{m+1}}{m+1} + C$ (where $m \neq -1$), we get:

This formula applies for $k > 1$. For $k=1$, it becomes $\int \frac{A_1}{x-a} dx = A_1 \ln|x-a| + C$.

Step 5: Combine the results. Summing the results from integrating all the partial fractions gives the final answer.

Key Takeaways:

• When a denominator has a repeated linear factor $(ax+b)^n$, the partial fraction decomposition must include terms $\frac{A_1}{ax+b}, \frac{A_2}{(ax+b)^2}, \dots, \frac{A_n}{(ax+b)^n}$.
• Integrals of the form $\int \frac{A}{(x-a)^k} dx$ for $k > 1$ are solved using the power rule, resulting in $\frac{-A}{(k-1)(x-a)^{k-1}}$.
• The integral of $\frac{A}{x-a}$ is $A \ln|x-a|$.
• Strategic substitution of roots and coefficient comparison are key methods for finding the unknown coefficients in the partial fraction decomposition.

Answer: The general approach involves setting up the correct partial fraction decomposition for repeated roots, solving for the coefficients, and then integrating each term using standard integration rules, including the power rule for negative exponents and the logarithm rule.