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Calculus AB · Unit 6 · 6.10
AP Calculus · Integration and Accumulation of Change
Integrating Using Long Division and Completing the Square
Neither is an integration technique. Both are algebra that turns an unfamiliar integrand into a familiar one.
key idea$\text{Rewrite the integrand into a form you already know how to integrate}$
The procedure
Neither of these is an integration technique — both are algebra applied before integrating. Long division handles top-heavy fractions; completing the square converts an awkward quadratic denominator into an $\arctan$ or $\arcsin$ form.
1
Diagnose which tool you need
Numerator degree $\ge$ denominator degree → long division. Irreducible quadratic in the denominator with no matching numerator → complete the square.
in the integrand out a choice
2
Long division: divide until the remainder is proper
$\dfrac{N}{D}=Q+\dfrac{R}{D}$ with $\deg R\lt\deg D$. The quotient $Q$ is a polynomial and integrates instantly.
out polynomial $+$ proper fraction
3
Completing the square: rewrite $ax^{2}+bx+c$
Force it into $(x+h)^{2}\pm k$. Then substitute $u=x+h$ so the denominator becomes $u^{2}+k$ — the arctangent form.
out a recognisable standard integral
4
Integrate each piece and undo any substitution
Polynomial terms by the power rule; the remaining fraction by log or arctangent.
out the answer in terms of $x$
| Integrand | Symptom | Tool | Becomes |
|---|
| $\dfrac{x^{2}}{x+1}$ | top degree $\ge$ bottom | long division | $x-1+\dfrac{1}{x+1}$ |
| $\dfrac{2x+3}{x+1}$ | equal degrees | long division | $2+\dfrac{1}{x+1}$ |
| $\dfrac{1}{x^{2}+4x+13}$ | irreducible quadratic | complete the square | $\dfrac{1}{(x+2)^{2}+9}$ |
| $\dfrac{1}{\sqrt{-x^{2}+4x}}$ | quadratic under a root | complete the square | $\dfrac{1}{\sqrt{4-(x-2)^{2}}}$ |
| $\dfrac{2x+4}{x^{2}+4x+13}$ | numerator $=$ derivative of denominator | substitution | $\ln\lvert x^{2}+4x+13\rvert$ |
The last row is the check that comes first: if the numerator is the denominator's derivative, use substitution and neither of these tools is needed.
Improper fractionNumerator degree $\ge$ denominator degree. Must be divided before anything else.
Irreducible quadraticA quadratic with negative discriminant — it never factors over the reals, so partial fractions cannot help.
Completing the squareRewriting $ax^{2}+bx+c$ as $(x+h)^{2}\pm k$, which exposes the arctangent or arcsine form.
The derivative checkIf the numerator equals the denominator's derivative, substitution finishes it in one step.
Worked examples
EX 1
Long division
$\int\dfrac{x^{2}}{x+1}\,dx$Evaluate $\displaystyle\int\dfrac{x^{2}}{x+1}\,dx$.
1
Diagnose
Numerator degree $2$, denominator degree $1$ — improper, so divide.
2
Divide
$x^{2}\div(x+1)=x-1$ remainder $1$, so $\dfrac{x^{2}}{x+1}=x-1+\dfrac{1}{x+1}$.
3
Verify the division
$(x-1)(x+1)+1=x^{2}-1+1=x^{2}$. ✓
4
Integrate term by term
$\dfrac{x^{2}}{2}-x+\ln\lvert x+1\rvert+C$.
5
Note the pay-off
Dividing turned one hard fraction into two elementary integrals.
EX 2
Completing the square
$\int\dfrac{dx}{x^{2}+4x+13}$Evaluate $\displaystyle\int\dfrac{dx}{x^{2}+4x+13}$.
1
Check for substitution first
The derivative of the denominator is $2x+4$, which does not appear in the numerator. So substitution alone will not work.
2
Complete the square
$x^{2}+4x+13=\left(x^{2}+4x+4\right)+9=(x+2)^{2}+9$.
3
Substitute
$u=x+2$, $du=dx$: the integral becomes $\displaystyle\int\dfrac{du}{u^{2}+9}$.
4
Recognise the arctangent form
$\displaystyle\int\dfrac{du}{u^{2}+a^{2}}=\dfrac1a\arctan\dfrac{u}{a}$, here with $a=3$.
5
Undo the substitution
$\dfrac13\arctan\dfrac{x+2}{3}+C$.
EX 3
Split the numerator
both tools at onceEvaluate $\displaystyle\int\dfrac{2x+7}{x^{2}+4x+13}\,dx$.
1
Compare with the denominator's derivative
$\dfrac{d}{dx}\left(x^{2}+4x+13\right)=2x+4$. The numerator is $2x+7=(2x+4)+3$.
2
Split accordingly
$\displaystyle\int\dfrac{2x+4}{x^{2}+4x+13}dx+\int\dfrac{3}{x^{2}+4x+13}dx$.
3
First piece by substitution
The numerator is exactly the derivative, so this gives $\ln\left(x^{2}+4x+13\right)$ — no absolute value needed, since the quadratic is always positive.
4
Second piece from Example 2
$3\cdot\dfrac13\arctan\dfrac{x+2}{3}=\arctan\dfrac{x+2}{3}$.
5
Combine
$\ln\left(x^{2}+4x+13\right)+\arctan\dfrac{x+2}{3}+C$. Splitting the numerator into ‘derivative part $+$ constant’ is the standard move for this shape.
EX 4
Under a square root
$\arcsin$ formEvaluate $\displaystyle\int\dfrac{dx}{\sqrt{-x^{2}+4x}}$.
1
Complete the square inside the root
$-x^{2}+4x=-\left(x^{2}-4x\right)=-\left[(x-2)^{2}-4\right]=4-(x-2)^{2}$.
2
Substitute
$u=x-2$: the integral becomes $\displaystyle\int\dfrac{du}{\sqrt{4-u^{2}}}$.
3
Recognise the arcsine form
$\displaystyle\int\dfrac{du}{\sqrt{a^{2}-u^{2}}}=\arcsin\dfrac{u}{a}$, with $a=2$.
4
Undo the substitution
$\arcsin\dfrac{x-2}{2}+C$.
5
Mind the leading minus
Factoring $-1$ out of $x^{2}-4x$ flips the signs inside the bracket. Losing that step gives $(x-2)^{2}-4$ and the wrong standard form entirely.
Where it goes wrong
Check for substitution before either tool
If the numerator is the denominator's derivative, one substitution finishes it. Long division and completing the square are for when that fails.
Long division only for improper fractions
If the numerator's degree is already lower, dividing achieves nothing. Compare degrees first.
Watch the leading minus when completing the square
Factoring $-1$ out reverses the signs inside the bracket — Example 4. This is where the arcsine form is most often lost.
$x^{2}+4x+13$ is always positive
Its discriminant is negative, so $\ln$ of it needs no absolute value. Check the discriminant before deciding.