How can completing the square uncover an inverse-tangent integral?
You will be able to: Complete a quadratic square and account for scale factors in integration.
How can completing the square uncover an inverse-tangent integral?
Rearranging puzzle pieces can reveal a familiar shape. Rewriting a quadratic as a square plus a positive constant can reveal the inverse-tangent derivative pattern.
A useful starting point: When should we divide before looking for an antiderivative? →
Words and symbols before equations
- Complete the square
- Rewrite x²+bx+c as (x+b/2)²+c−b²/4.
- Shift
- A replacement such as u=x+1.
- Scale a
- A positive constant in a²+u².
- arctan
- The inverse tangent; its derivative is 1/(1+u²).
What this picture assumes
Original model; readouts are rounded. f=1/((x+1)²+a²); a>0. F=arctan((x+1)/a)/a. Completing a square identifies the shift −1 and positive scale a. All x are allowed for a>0.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- a=2, x=0; f=0.2; F=0.231824. F′=[1/a²]/[1+((x+1)/a)²]=f. a stays positive.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
The denominator x²+2x+5 equals (x+1)²+4. Set u=(x+1)/2, so dx=2du and the denominator is 4(u²+1).
Therefore ∫dx/(x²+2x+5)=(1/2)∫du/(1+u²)=(1/2)arctan((x+1)/2)+C.
In general, for a>0, ∫du/(a²+u²)=(1/a)arctan(u/a)+C. Differentiating verifies both the outside factor and the inside scaling.
A logarithm would need a numerator proportional to the denominator’s derivative. A constant numerator over a completed positive square generally fits inverse tangent instead. A quadratic with real zeros needs separate domain care.
A worked example, step by step
Evaluate ∫₀¹ dx/(x²+2x+2).
- Complete the square: x²+2x+2=(x+1)²+1.
- An antiderivative is arctan(x+1), since the inner derivative is 1.
- Evaluate at both endpoints: arctan2−arctan1=arctan2−π/4.
- The value is approximately 0.322 radians; it is positive because the integrand is positive throughout.
Completing the square must preserve the constant: (x+1)² already includes +1. Account for that before applying a standard formula.
Is x²+2x+5 equal to (x+1)²+5?
Compare with an explanation
No. That would add an extra 1; the correct remaining constant is 4.
Predict. Change one thing. Explain.
Change the positive constant a in 1/((x+1)²+a²). Compare the peak height and the antiderivative slope at a chosen x. Explain why the factor 1/a is necessary.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
a=2, x=0; f=0.2; F=0.231824. F′=[1/a²]/[1+((x+1)/a)²]=f. a stays positive.
Original model; readouts are rounded. f=1/((x+1)²+a²); a>0. F=arctan((x+1)/a)/a. Completing a square identifies the shift −1 and positive scale a. All x are allowed for a>0.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the rate units, signed contributions, theorem conditions, domain or antiderivative check. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionIntegrate 1/(x²−6x+13). Complete the square and verify the result.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: The denominator is (x−3)²+4.
- 1 point: Use u=(x−3)/2, so dx=2du.
- 1 point: The result is (1/2)arctan((x−3)/2)+C.
- 1 point: Its derivative is (1/4)/(1+(x−3)²/4)=1/((x−3)²+4), valid for all real x.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What pattern follows a positive completed square?
The inverse-tangent derivative form.
RECALL 2Why does scale matter twice?
The outside coefficient and inside chain-rule factor both affect the derivative.
RECALL 3How can a mistaken constant be caught?
Expand the completed square before integrating.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How can completing the square uncover an inverse-tangent integral?
- x²+bx+c=(x+b/2)²+c−b²/4.
- ∫du/(a²+u²)=(1/a)arctan(u/a)+C, a>0.
- Verify all scale factors by differentiation.
Remember: Completing the square must preserve the constant: (x+1)² already includes +1. Account for that before applying a standard formula.
Conditions: Original model; readouts are rounded. f=1/((x+1)²+a²); a>0. F=arctan((x+1)/a)/a. Completing a square identifies the shift −1 and positive scale a. All x are allowed for a>0.
Refresh Kid · AP Calculus AB Unit 6 · Objectives FUN-6.D · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 6.10, FUN-6.D. CED effective Fall 2020 and Fall 2026 clarifications checked September 17, 2026. AB scope includes 6.1–6.10 and 6.14. Topics 6.11–6.13 are BC-only; the numbering gap is intentional. Topic 6.14 consolidates the AB antidifferentiation objectives and Skill 1.C. Focused lesson titles, examples and questions are original Refresh Kid material.
Distinguish signed accumulation, unsigned area and initial amount. Error direction needs interval-wide shape information. FTC hypotheses, variable-bound chain factors, integration constants, transformed bounds and domain restrictions are explicit. Bounded jumps are treated separately from differentiability of accumulation. No improper-integral evaluation, integration by parts or partial-fraction decomposition is taught in this AB unit.
The Organic Chemistry Tutor Fundamental Theorem of Calculus Part 1 video title, creator and description were checked; the full video was not reviewed. Khan Academy’s unit destination was checked, but its JavaScript lesson contents were not fully readable by the research tool. OpenStax Sections 5.2, 5.3 and 5.5 were consulted for conceptual cross-checking. No provider scripts, questions, diagrams or artwork were copied. Refresh Kid is not affiliated with or endorsed by these providers.
GitHub’s 3D website collection informed optional spatial inspection and camera controls. The original tank geometry uses existing self-hosted Three.js with its MIT license. Its 2×2 dm base and water depth use the same volume as the rate calculation: 1 dm³=1 L. The initial water is blue and newly accumulated inflow is teal. The rate graph is an abstract amount calculation; it is not the physical shape of water. Camera rotation changes only the view. Full 2D graphs, readouts and explanations remain available without WebGL.
Independent teacher review and observation of students remain pending. Technical checks do not certify mathematical accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Calculus AB questions and scoring guides. The archive spans multiple units; no entire exam question is assigned here.
Learn → Explore → Practice → Review is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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