Long Division: How to Deconstruct complex Integrals

Using long division in calculus involves dividing the polynomial numerator by the polynomial denominator to separate the integral into more manageable parts. This division yields a simpler polynomial plus a fraction where the numerator’s degree is less than the denominator’s. The reduced fraction often requires substitution or partial fractions to find the integral.

In calculus, when faced with integrating a complex rational function \( \frac{P(x)}{Q(x)} \) where \( P(x) \) and \( Q(x) \) are polynomials, long division is a useful technique. It simplifies the function by dividing \( P(x) \) by \( Q(x) \), yielding a quotient that is a simpler polynomial plus a remainder fraction, where the numerator’s degree is less than the denominator’s.

Long division in calculus divides \( P(x) \), a higher degree polynomial, by \( Q(x) \), to simplify \( \int \frac{P(x)}{Q(x)} \, dx \). The result is a quotient polynomial plus a remainder fraction \( \frac{R(x)}{Q(x)} \), where \( \deg(R) < \deg(Q) \). Integrating the quotient is straightforward. For \( \frac{R(x)}{Q(x)} \).

Let’s consider an integral involving a rational function where we apply the long division method. Suppose we want to integrate:

\(\int \frac{3x^3 + 2x^2 – 5x + 4}{x^2 – 2x + 1} \, dx\)

1. Long Division:

   First, we perform long division on \( \frac{3x^3 + 2x^2 – 5x + 4}{x^2 – 2x + 1} \).

   Result: \( 3x + 8 + \frac{11x + 12}{x^2 – 2x + 1} \)

2. Integrate the Polynomial and Fraction:

   \(  \int (3x + 8) \, dx + \int \frac{11x + 12}{x^2 – 2x + 1} \, dx   \)

   The first part, \( \int (3x + 8) \, dx \), integrates to:

     \(     \frac{3}{2}x^2 + 8x + C_1     \)

   For the fractional part, notice \( x^2 – 2x + 1 = (x-1)^2 \). The integral simplifies to:

     \(     \int \frac{11x + 12}{(x-1)^2} \, dx     \)

  We can simplify this by setting \( u = x – 1 \), then \( du = dx \) and \( x = u + 1 \). Substituting these gives:

     \(     \int \frac{11(u+1) + 12}{u^2} \, du = \int \frac{11u + 23}{u^2} \, du     \)

  Separate the terms and integrate:

     \(     11 \int \frac{u}{u^2} \, du + 23 \int \frac{1}{u^2} \, du = 11 \int u^{-1} \, du + 23 \int u^{-2} \, du     \)

     Integrating these yields:

     \(     11 \ln |u| – \frac{23}{u} + C_2     \)

     Substitute back for \( x \):

     \(     11 \ln |x-1| – \frac{23}{x-1} + C_2     \)

3. Combine Results:

   The complete integral is:

   \(   \frac{3}{2}x^2 + 8x + 11 \ln |x-1| – \frac{23}{x-1} + C   \)

Here, \( C \) represents the constant of integration.