Differentiation has one algorithm that works on everything put in front of it. Integration doesn't. The gap between "this integral looks hard" and "this integral is easy" is very often pure algebra, done BEFORE any integration rule gets touched. Two moves carry disproportionate weight. One is polynomial long division on a rational expression whose numerator's degree isn't smaller than its denominator's. The other is completing the square to expose a u^2\pm a^2 shape hiding inside a quadratic.
Neither move is calculus at all. They're the same algebra from College Algebra, redeployed. The skill this section actually teaches is noticing WHEN to reach for them. An integrand that looks unfamiliar is often just one long-division or one completing-the-square step away from something you've known how to integrate all along. Getting stuck on an integral is very often a signal to stop reaching for a new integration rule and check whether the algebra underneath it has been reshaped yet.
Two algebra moves do disproportionate work before any integration rule ever gets touched. Polynomial long division divides a rational expression's numerator by its denominator when the numerator's degree isn't smaller. Completing the square rewrites a quadratic as a single squared term plus a constant. \int \frac{x^2}{x+1}\,dx looks like it needs partial fractions two chapters from now. But the numerator's degree isn't less than the denominator's, so long division applies FIRST: \frac{x^2}{x+1} = x - 1 + \frac{1}{x+1}. That's three pieces the basic rules already handle. Completing the square plays the same "reshape before integrating" role for \int \frac{dx}{x^2+6x+13}. Rewriting the denominator as (x+3)^2+4 turns an opaque quadratic into the u^2+a^2 shape an inverse-tangent antiderivative recognizes, later in this chapter. Neither move changes what the integrand actually IS. Only the form it's written in changes, which is the whole point.
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
Factor 2x^2 - 8 completely. Type the result (using parentheses and * for multiplication).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
GCF of 2x^2 and 8 is 2: 2x^2-8=2(x^2-4). Then x^2-4=(x+2)(x-2), giving 2(x+2)(x-2).
Correct answer
Also accepted:
Why
Pull out the greatest common factor first, then check whether what's left factors further.
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
Using the quadratic formula on x^2 + 2x - 8 = 0, the solutions are:
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Signs flipped — with b=2, the formula's -b is -2, and the discriminant is 4+32=36, giving x = \frac{-2\pm6}{2}.
Both signs are flipped from the correct roots — the correct solutions are x=2 and x=-4, not x=4 and x=-2. Using the formula, x=\frac{-2+6}{2}=2, not -2.
Check the discriminant: b^2-4ac = 4+32=36, so \sqrt{36}=6, not something that produces x=1.
Why
With a=1, b=2, c=-8: discriminant = 4+32=36, so x = \frac{-2 \pm 6}{2}, giving x=2 or x=-4. Check x=2: 4+4-8=0 ✓.
Identify a, b, c carefully, then compute the discriminant b^2-4ac before applying the formula.
Reshape a rational expression before integrating
Drag the sliders -- notice the asymptote and hole structure change with the coefficients, the exact structure long division needs to account for.
Loading interactive…
Completing the square exposes an arctan shape
Slide h — the antiderivative shape only becomes recognizable once the denominator is rewritten as (x-h)^2+a^2.
Loading interactive…
Worked example
Long division before integrating
-
1
Evaluate \displaystyle\int \frac{x^2+1}{x-1}\,dx.
-
2
The numerator's degree is not less than the denominator's, so divide first: x^2+1 = (x-1)(x+1) + 2, giving \frac{x^2+1}{x-1} = x+1+\frac{2}{x-1}.
-
3
Integrate term by term: \int(x+1)\,dx = \frac{x^2}{2}+x, and \int\frac{2}{x-1}\,dx = 2\ln|x-1|.
-
4
\boxed{\int\frac{x^2+1}{x-1}\,dx = \frac{x^2}{2}+x+2\ln|x-1|+C}
Loading interactive…
Where this goes next
Partial fractions, several chapters ahead
Long division here is a preview of a bigger idea. Partial fractions splits an improper OR proper rational function into simpler pieces before integrating. This section's division step is the move that handles the improper case first.
Check your understanding (counts toward your progress)
Complete the square: x^2-10x-22=(x-5)^2+k. Find k.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
h=-5, so k=-22-(-5)^2\approx-47.0 -- this rewrites the denominator into a form ready for a trig or inverse-trig substitution.
Correct answer
Also accepted:
Why
h=\dfrac{b}{2}=-5; k=c-h^2.
Before integrating \displaystyle\int \dfrac{x^3}{x-2}\,dx, what should be done first?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
A plain substitution doesn't clear the mismatch in degree between numerator and denominator here.
Partial fractions applies to a PROPER rational function — this one isn't proper yet.
The expression isn't in a directly-integrable form until the division is done.
Why
The numerator's degree (3) is not less than the denominator's (1), so long division comes first.
Compare the degree of the numerator to the degree of the denominator.
Before integrating \displaystyle\int\dfrac{-3x^2+3x+6}{x-8}\,dx, perform polynomial long division: \dfrac{-3x^2+3x+6}{x-8}=-3x-21+\dfrac{R}{x-8}. Find the remainder R.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
R=-3(8)^2+3(8)+6=-162.
Correct answer
Also accepted:
Why
By the Remainder Theorem, R equals the numerator evaluated at x=8.