Can I Integrate by Parts with a Rational?

Introduction
Integration by parts is a fundamental technique in calculus that allows us to integrate products of functions. However, it is not immediately clear whether this technique can be applied to rational functions, which involve quotients of polynomials. This article explores the conditions under which integration by parts can be used with rational functions and provides step-by-step instructions for performing the integration process.

Conditions for Integration by Parts
The integration by parts formula states that for two functions $u$ and $v$, the integral of $uv$ is given by:

$$ \int uv dx = u \int v dx – \int \left( \frac{du}{dx} \int v dx \right) dx $$

In order for this formula to be valid, both $u$ and $v$ must have continuous derivatives. This means that rational functions can only be integrated by parts if the denominator polynomial has no real roots.

Step-by-Step Integration
If the conditions stated above are met, we can integrate a rational function by parts using the following steps:

1. Decompose the rational function into partial fractions. This involves expressing the function as a sum of simpler fractions with linear or quadratic denominators.
2. Choose $u$ and $dv$ judiciously. In general, we want to choose $u$ to be the term with the highest derivative and $dv$ to be the term with the lowest derivative.
3. Use the integration by parts formula to integrate $uv$. This will result in an expression involving integrals of $u$ and $v$.
4. Solve for the integrals of $u$ and $v$. This may require further integration by parts or other techniques.
5. Combine the integrals to obtain the final result.

Examples

Example 1:

Integrate the rational function:

$$ \int \frac{x^2 + 1}{x^2 – 1} dx $$

Solution:

1. Decompose the function into partial fractions:

$$ \frac{x^2 + 1}{x^2 – 1} = 1 + \frac{2}{x^2 – 1} $$

2. Choose $u = 1$ and $dv = \frac{1}{x^2 – 1} dx$.

3. Use integration by parts:

$$ \int \left( 1 \cdot \frac{1}{x^2 – 1} \right) dx = x + \int \frac{2x}{(x^2 – 1)^2} dx $$

4. Solve for the integral of $v = \frac{1}{x^2 – 1}$:

$$ \int \frac{1}{x^2 – 1} dx = \frac{1}{2} \ln |x^2 – 1| + C $$

5. Combine the integrals:

$$ \int \frac{x^2 + 1}{x^2 – 1} dx = x + \frac{2x}{(x^2 – 1)^2} \cdot \frac{1}{2} \ln |x^2 – 1| + C $$

Applications

Integration by parts with rational functions has applications in various fields, including:

• Physics: Calculating forces and moments in mechanics
• Engineering: Designing structures and components
• Economics: Modeling supply and demand, and other economic phenomena
• Finance: Valuing options and other financial instruments

Conclusion

Integration by parts can be used with rational functions if the denominator polynomial has no real roots. By decomposing the rational function into partial fractions and carefully choosing $u$ and $dv$, we can apply the integration by parts formula to obtain the desired integral. This technique has widespread applications in science, engineering, economics, and finance.