Long Division: How to Deconstruct complex Integrals
The long division method in calculus simplifies rational functions before integration. When the numerator’s degree is higher than the denominator’s, long division helps rewrite the function into a polynomial plus a remainder over the original denominator. This transformation makes integration more straightforward, allowing for the application of basic integration rules to the polynomial and the proper treatment of the fractional remainder.
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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.
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