Translation Not Available Yet
Integration by Completing the Square is not yet available in Deutsch. Showing the English version below. All formulas and calculations work the same.
Zuletzt aktualisiert: 31. Juli 2026
Integration by Completing the Square
Ersteller
Dharmendra SinghPrüfer
Ersteller
Dharmendra SinghPrüfer
Ersteller
Dharmendra SinghPrüfer
Ersteller
Dharmendra SinghPrüfer
Formel
For ∫ dx/(ax²+bx+c), write ax²+bx+c = a[(x-h)²+k] with h = -b/(2a), k = (4ac-b²)/(4a²), and D = b²-4ac.
Wobei:
- a=Quadratic coefficient
- b=Linear coefficient
- c=Constant term
- h=Horizontal shift of the completed square
- k=Constant inside the completed-square form
- D=Discriminant that selects the integration case
Rechenbeispiele
Complex-root denominator
For x²+2x+5, the completed square reveals a positive shifted constant.
- 1Compute h = -b/(2a) = -2/2 = -1.
- 2Compute k = (4ac-b²)/(4a²) = (20-4)/4 = 4.
- 3Compute D = b²-4ac = 4-20 = -16, so the antiderivative uses an arctan form.
Repeated-root denominator
A perfect-square denominator leads to the repeated-root case.
- 1Compute h = -(-4)/(2·1) = 2.
- 2Compute k = (16-16)/4 = 0.
- 3Since D = (-4)²-4·1·4 = 0, the denominator has a repeated root.
Two-real-root denominator
When the discriminant is positive, the quadratic factors over the reals.
- 1Compute h = 0 because b = 0.
- 2Compute k = (4·1·(-1)-0)/(4·1²) = -1.
- 3Compute D = 0-4·1·(-1) = 4, so the denominator has two real roots.
Negative leading coefficient
The same algebra works even when a is negative.
- 1Compute h = -6/(2·-2) = 1.5.
- 2Compute k = (4·-2·1-36)/(4·4) = -2.75.
- 3Compute D = 36-4·-2·1 = 44, so the denominator corresponds to two real roots.
Einführung
This calculator does not directly output the full antiderivative. Instead, it gives the completed-square parameters that tell you which integration strategy applies to ∫dx/(ax²+bx+c). That makes it useful for checking algebra, planning substitutions, and understanding why a quadratic denominator turns into an arctan expression, a repeated-root reciprocal, or a real-root partial-fractions problem.
What this calculator shows
The tool extracts the quadratic structure behind the integral by computing h, k, and the discriminant. Those three values summarize the geometry of the denominator and indicate the most natural antiderivative form.
h tells you where the parabola is horizontally centered.
k tells you the constant left after completing the square.
The discriminant D decides whether the quadratic has complex roots, a double root, or two distinct real roots.
How completing the square works
Starting with ax²+bx+c, factor out a from the quadratic terms and rewrite the bracket as a perfect square plus a constant. This isolates the shifted variable x-h and prepares the denominator for substitution.
Factor a: ax²+bx+c = a[x²+(b/a)x+c/a].
Add and subtract (b/2a)² inside the bracket.
Rewrite the result as a[(x-h)²+k].
If you make an algebra slip while completing the square by hand, the values of h and k from this tool can help you catch it immediately.
Why the discriminant matters
The discriminant D = b²-4ac classifies the quadratic. That classification determines the integration method you usually use next.
- D < 0:
no real roots, so the completed-square denominator stays positive or negative and often leads to arctan.
- D = 0:
repeated root, giving a reciprocal linear-square structure.
- D > 0:
two real roots, so partial fractions or root-based rewriting becomes natural.
The calculator reports the case type only; it does not expand the full symbolic antiderivative.
Interpreting the arctan case
When D < 0, the quadratic cannot be factored over the reals. After completing the square, the denominator becomes a multiple of (x-h)²+k with k > 0, which matches the standard ∫du/(u²+a²) pattern.
Set u = x-h.
Factor constants outside the integral.
Use the identity ∫du/(u²+a²) = (1/a) arctan(u/a) + C.
Interpreting the repeated-root case
When D = 0, the quadratic is a perfect square. The integral reduces to the reciprocal of a squared linear expression after a simple shift.
The denominator becomes a(x-h)².
Use the substitution u = x-h.
Apply ∫u⁻² du = -u⁻¹ + C.
Interpreting the two-real-roots case
When D > 0, the denominator factors into two linear terms over the reals. Completing the square still gives insight, but many students then switch to factoring and partial fractions to integrate.
Find the real roots if needed.
Write the denominator as a(x-r₁)(x-r₂).
Decompose into simpler fractions before integrating.
Input rules and validation
The coefficient a must be nonzero because the technique is specifically about quadratic denominators. All three coefficients must be finite numbers.
Use decimals or integers.
Avoid leaving fields blank.
If a = 0, the denominator is linear rather than quadratic.
FAQ
What does h represent?
h is the horizontal shift of the parabola after completing the square. It is the value that centers the quadratic around x = h.
What does k represent?
k is the constant term left inside the completed-square expression a[(x-h)²+k]. It helps determine whether the denominator matches an arctan pattern.
Why does the calculator show a case type instead of the full integral?
The goal is to classify the quadratic denominator and provide the structural values used in manual integration. That makes it easier to study the method and verify algebraic steps.
Why must a be nonzero?
If a were zero, the denominator would be linear rather than quadratic, so completing the square would not be the relevant technique.
When does the arctan form appear?
It appears when the discriminant is negative, which means the quadratic has no real roots and can be rewritten as a positive square plus a positive constant.
What happens when the discriminant is zero?
The quadratic has a repeated real root and becomes a perfect square. The integral usually simplifies to a reciprocal linear expression after substitution.
What happens when the discriminant is positive?
The quadratic has two distinct real roots. In many cases, partial fractions become the most convenient next step.
Can I use decimal coefficients?
Yes. The calculator accepts finite decimal values for a, b, and c and rounds reported results to six decimal places.
Does this work for definite integrals?
Yes for the algebraic setup. You can use h, k, and the case type to transform the integrand before applying limits.