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6.10 Integrating Functions Using Long Division and Completing the Square

Syllabus
2020
Topic
6.10
Level
—

Learning objectives

—FUN-6.D—For integrands requiring substitution or rearrangements into equivalent forms: (a) Determine indefinite integrals. (b)…—Topic 6.10• FUN-6.D For integrands requiring substitution or rearrangements into equivalent forms: (a) Determine indefinite integrals. (b) Evaluate definite integrals.• FUN-6.D.3 Techniques for finding antiderivatives include rearrangements into equivalent forms, such as long division and completing the square.• Enduring understanding FUN-6: Recognizing opportunities to apply knowledge of geometry and mathematical rules can simplify integration.

Rewrite First, Then Integrate

Before choosing a new integration rule, check whether algebra can expose a familiar one. Polynomial long division separates a polynomial part from a proper rational remainder; completing the square converts a quadratic into a shifted square plus a constant.

Integrand feature Rewrite Structure revealed
numerator degree ≥\ge denominator degree polynomial long division polynomial terms plus a proper fraction
quadratic x2+bx+cx^2+bx+c (x+b/2)2+c−b2/4(x+b/2)^2+c-b^2/4 shifted u2+a2u^2+a^2 or related form

For x≠−1x\ne-1, x2+1x+1=x−1+2x+1\frac{x^2+1}{x+1}=x-1+\frac{2}{x+1}. Therefore ∫x2+1x+1 dx=x22−x+2ln⁡∣x+1∣+C\int\frac{x^2+1}{x+1}\,dx=\frac{x^2}{2}-x+2\ln|x+1|+C. Multiplying the quotient by the divisor and adding the remainder verifies the rewrite.

Since x2+4x+8=(x+2)2+4x^2+4x+8=(x+2)^2+4, ∫−20dxx2+4x+8=[12arctan⁡(x+22)]−20=12(π/4−0)=π/8\int_{-2}^{0}\frac{dx}{x^2+4x+8}=\left[\frac12\arctan\left(\frac{x+2}{2}\right)\right]_{-2}^{0}=\frac12(\pi/4-0)=\pi/8. The completed square identifies the inverse-tangent form.

A rewrite must be algebraically equivalent on the original domain. Long division does not restore a denominator's excluded zeros, and completing the square does not change definite-integral bounds unless a separate variable substitution is introduced.