Integrating Functions Using Long Division and Completing the Square
Introduced| English | Meaning |
|---|---|
| completing the square/kəmˈpliːtɪŋ ðə skweə/ | Rewriting a quadratic expression as a squared linear expression plus or minus a constant. |
| polynomial long division/ˌpɒlɪˈnəʊmɪəl lɒŋ dɪˈvɪʒn/ | A step-by-step method for dividing one polynomial by another to obtain a quotient and remainder. |
Rewrite first, then integrate
- Some rational functions don't match any basic antiderivative — until you rewrite them.
- Two algebra tools prepare a stubborn integrand: long division and completing the square.
- Neither is calculus; both just reshape the fraction into a friendlier form.
- Then a basic rule or a known integral finishes the job.
Long division for "top-heavy" fractions
- If the numerator's degree is $\ge$ the denominator's, the fraction is improper.
- Use polynomial long division 多项式长除法 to split it into a polynomial plus a proper remainder.
- $\dfrac{x^2}{x+1}=x-1+\dfrac{1}{x+1}$ — now each piece integrates easily.
- $\displaystyle\int\dfrac{x^2}{x+1}\,dx=\dfrac{x^2}{2}-x+\ln|x+1|+C$.
A rational curve to rewrite
y = a / (x − b)
An improper fraction like $\tfrac{x^2}{x+1}$ becomes a line plus a small remainder — each piece integrates easily.
Use polynomial long division when the numerator's degree is...
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An improper fraction (top-heavy) calls for division.
Long division gives $\dfrac{x^2}{x+1}=$
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$x^2=(x+1)(x-1)+1$, so the quotient is $x-1$ remainder $1$.
Completing the square for inverse-trig forms
- A denominator like $x^2+4x+5$ hides a perfect square: complete the square 配方 to $(x+2)^2+1$.
- That $\dfrac{1}{(x+2)^2+1}$ shape matches the arctangent antiderivative.
- $\displaystyle\int\dfrac{1}{x^2+4x+5}\,dx=\int\dfrac{1}{(x+2)^2+1}\,dx=\arctan(x+2)+C$.
- Completing the square turns a messy quadratic denominator into the clean $u^2+a^2$ pattern.
Completing the square, $x^2+4x+5=$
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$(x+2)^2=x^2+4x+4$, so $x^2+4x+5=(x+2)^2+1$.
The form $\dfrac{1}{(x+2)^2+1}$ integrates to an ____ function.
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$\int\tfrac{1}{u^2+1}\,du=\arctan u+C$.
Recognize the signal
- Numerator degree $\ge$ denominator degree → reach for long division first.
- An irreducible quadratic denominator (no real roots) → try completing the square toward arctan.
- These are setup moves; the actual integration is a basic rule or u-sub afterward.
- Spotting which rewrite to use is half the battle.
$\displaystyle\int\dfrac{x^2}{x+1}\,dx=$
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Divide first, then integrate $x-1+\tfrac1{x+1}$.
Match the signal to the rewrite: select all true pairings.
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Long division is for improper fractions only; both are pre-integration rewrites.
Do the long division only when the fraction is improper (numerator degree $\ge$ denominator's) — dividing a proper fraction gets you nowhere. And after completing the square, keep the constant right: $x^2+4x+5=(x+2)^2+1$, not $(x+2)^2+5$. A wrong constant breaks the arctan match.
Integrate $\displaystyle\int\dfrac{x^2}{x+1}\,dx$.
- Long division: $\dfrac{x^2}{x+1}=x-1+\dfrac{1}{x+1}$.
- Integrate term by term: $\dfrac{x^2}{2}-x+\ln|x+1|+C$.
- (Each rewritten piece uses a basic antiderivative rule.)
When a rational integrand fits no basic rule, rewrite it first: use polynomial long division on an improper fraction (numerator degree $\ge$ denominator's), or complete the square in an irreducible quadratic denominator to reach an arctangent form. Then finish with a basic antiderivative.