Mastering Completing the Square Integration

An integral with a quadratic denominator, like ∫ dx/(x² + 4x + 13), blocks the standard arctan or arcsin formulas until you rewrite that quadratic. Completing the square integration solves this: it turns the quadratic into (x + p)² + q², which matches the form 1/(u² + a²) or 1/√(a² − u²). You then substitute u = x + p and apply the arctangent or arcsine rule directly. OpenStax Calculus Vol. 2 covers this in section 3.3 before trigonometric substitution. The rewrite adds (b/2)² and subtracts it, adjusting the constant, and takes under a minute with practice.

Complete the square when the integrand contains a quadratic that does not factor into real linear terms, especially in a denominator or inside a square root. The most common errors happen when a ≠ 1: factor out the leading coefficient a before adding (b/2)² inside the parentheses, or the constant term will be wrong. For an integral like ∫ dx/(3x² + 6x + 10), factor the 3 out first: 3(x² + 2x) + 10, then complete the square on x² + 2x. The vertex form a(x − h)² + k with h = −b/(2a) and k = c − b²/(4a) is the target. Use this technique when the discriminant b² − 4ac is negative, so the quadratic is irreducible over the reals.

The integral ∫ du/(u² + a²) equals (1/a) arctan(u/a) + C, a result listed in Appendix A of OpenStax Calculus Vol. 2. After completing the square, set u equal to the linear term (x + b/2) and a equal to the square root of the adjusted constant. For ∫ dx/(x² + 4x + 13), the result is (1/3) arctan((x + 2)/3) + C. The failure case: if 4c − b² is not positive, the constant factor becomes imaginary, and you need a different method or partial fractions with complex numbers. Check 4c − b² > 0 before applying the arctangent formula.

The integral ∫ du/√(a² − u²) equals arcsin(u/a) + C, also from the integral table. This pattern appears when the quadratic appears under a square root after completing the square and the coefficient of the squared term is negative. For ∫ dx/√(x² + 2x + 5), the result is arcsinh((x + 1)/2) + C (the hyperbolic arcsine form, since a² − u² requires a negative sign before u²). A common mistake: applying the arcsine formula to a form that should use arctan. Test whether the integrand has √(something) in the denominator and whether the squared term inside the root is negative. If it is negative, use arctan; if positive, use arctan or a hyperbolic substitution.

Example 1: ∫ dx/(x² − 6x + 10)

Complete the square: x² − 6x + 10 = (x − 3)² + 1. The integral becomes ∫ dx/((x − 3)² + 1). Substitute u = x − 3, du = dx. Result: ∫ du/(u² + 1) = arctan(u) + C = arctan(x − 3) + C. OpenStax Vol. 2 section 3.3 confirms this.

Example 2: ∫ dx/√(x² + 6x + 10)

Complete the square: x² + 6x + 10 = (x + 3)² + 1. The integral is ∫ dx/√((x + 3)² + 1). This matches the form for arcsinh: the result arcsinh(x + 3) + C.

Example 3: ∫ dx/(x² + 2x + 2)

Complete the square: x² + 2x + 2 = (x + 1)² + 1. The integral equals ∫ du/(u² + 1) with u = x + 1, giving arctan(x + 1) + C.

Failure case: odd b coefficient

For x² + 3x + 2, the square is (x + 3/2)² − 1/4. The integral becomes ∫ dx/((x + 3/2)² − 1/4). Here 4c − b² = 4(2) − 9 = −1, which is negative. The arctan form does not apply directly; use partial fractions or a different substitution. This is the most common failure mode for students.

Completing the square is the first step before trigonometric substitution, as section 3.3 of OpenStax shows. After rewriting the quadratic as a(x − h)² + k, substitute a trigonometric function: for √(a² − u²), use u = a sin θ; for √(a² + u²), use u = a tan θ. Partial fractions handle the case where the quadratic factors over the reals, when the discriminant is non-negative, but completing the square is faster for irreducible quadratics. If the denominator has repeated factors or higher-degree polynomials, combine completing the square with partial fraction decomposition.

Use this table to match the completed square form to the correct integral formula. The values come from the integral table in Appendix A of OpenStax Calculus Vol. 2.

Denominator Form	Completed Square	Substitution	Result
1/(x² + bx + c)	(x + p)² + q²	u = x + p, a = q	(1/q) arctan(u/q) + C
1/√(x² + bx + c)	(x + p)² + q²	u = x + p, a = q	arcsinh(u/q) + C
1/√(a² − (x + p)²)	(x + p)² = a² − u²	u = x + p, a constant	arcsin(u/a) + C
1/((x + p)² − q²)	(x + p)² − q²	u = x + p, a = q	(1/(2q)) ln|(u−q)/(u+q)| + C or arctanh

This technique is for first-year calculus students who need to evaluate integrals with quadratic denominators, and for tutors who want a textbook-aligned reference. It suits students who have already learned the basic arctan and arcsin formulas and need to handle quadratics that do not fit those forms directly. Skip this section if you need a refresher on how to complete the square step by step; that is a separate topic. Also skip if your integral has a cubic denominator or uses partial fractions with linear factors only; those are different methods. The one thing that most often goes wrong: forgetting to factor out the leading coefficient when a ≠ 1, which shifts the vertex's y-coordinate and produces a wrong constant under the arctan.