Complete the square before integrating

Try

Consider the integral of 1/(x² − 4x + 13). The denominator does not factor into real linear factors, and its derivative 2x − 4 is not the numerator. A logarithm substitution therefore does not immediately fit. Instead, rewrite the quadratic to expose a translated square:

x² − 4x + 13 = (x − 2)² + 9.

Check this identity by expanding. Completing the square has preserved the entire denominator, including its constant term; replacing 13 with 4 would change the function. Predict which familiar derivative involves a positive constant plus a square in its denominator. The inverse tangent derivative, d(arctan u)/du = 1/(1 + u²), supplies the relevant pattern.

Predict

Compare these three proposed antiderivatives before differentiating them:

Candidate	Proposed expression
A	arctan(x − 2)
B	arctan((x − 2)/3)
C	(1/3)arctan((x − 2)/3)

All three have an inverse tangent and a horizontal shift. That resemblance alone cannot choose the correct one. Predict what happens when the inner input is divided by 3: its derivative contributes a factor 1/3, while normalizing the quadratic denominator contributes a factor 9. Track both factors rather than memorizing the final coefficient.

Experiment

Make the change of variable u = (x − 2)/3. Then x − 2 = 3u, dx = 3 du, and the denominator is 9(1 + u²). Write the transformed numerator and denominator on separate lines:

∫ dx/((x − 2)² + 9) = ∫ 3 du/[9(1 + u²)] = (1/3)arctan u + C.

Thus candidate C is correct. Verify it independently by differentiation:

F′(x) = (1/3) · [1/(1 + ((x − 2)/3)²)] · (1/3) = 1/((x − 2)² + 9).

Now differentiate candidate B. Its derivative is 3/((x − 2)² + 9), three times the desired integrand. Candidate A instead has denominator 1 + (x − 2)², so changing only an outside coefficient cannot repair its incorrect horizontal scale. Write a one-sentence diagnosis of each failure.

Understand

For a definite integral from x = 2 to x = 5, transform the bounds along with the variable. The lower bound becomes u = 0; the upper becomes u = 1. Consequently the integral is (1/3)[arctan(1) − arctan(0)] = π/12. Inverse-trigonometric values are in radians in these differentiation formulas. A degree-mode calculator result cannot be substituted into the same formula without accounting for its unit conversion.

Original input x	Normalized input u	Principal arctangent
2	0	0
5	1	π/4

You can either evaluate the original-variable antiderivative at x = 2 and 5, or evaluate the transformed integral at u = 0 and 1. Mixing the original bounds with the u-variable integrand is an error. The integrand is positive throughout this interval, so the answer must be positive; reversing the limits would change its sign.

More generally, with a positive constant a, the integral of 1/((x − h)² + a²) is (1/a)arctan((x − h)/a) + C. Here h locates the shift and a is the positive square root of the constant added to the square. The substitution x − h = au explains the coefficient: dx supplies a, while the denominator supplies a². The integrand is defined for every real x because its denominator is at least a², which is positive.

This domain argument is part of method selection. A denominator such as (x − h)² − a² has real zeros and a different antiderivative structure; it cannot be handled by copying the plus-sign formula. If completing the square leaves zero rather than a positive constant, a reciprocal-square power rule applies away from the excluded input. The appearance of a quadratic alone does not guarantee an inverse-tangent answer.

Apply

For a new positive quadratic denominator, perform four checks: expand the completed square, identify the positive scale, differentiate the antiderivative, and check the original domain. Retain the leading coefficient if the quadratic is not monic. For example, factoring a nonzero constant outside a denominator changes the reciprocal integrand by the reciprocal constant, not by that same constant.

A nearby numerator can call for two mechanisms. Since x = (x − 2) + 2, the integral of x/((x − 2)² + 9) splits into a logarithmic part and twice the inverse-tangent part. An antiderivative is (1/2)ln((x − 2)² + 9) + (2/3)arctan((x − 2)/3). Differentiate both terms to recover (x − 2 + 2)/((x − 2)² + 9). The logarithm’s argument is always positive here; in other substitutions, any needed absolute value and excluded inputs must be retained.

Use a calculator only after obtaining the structure when an approximation is requested. Report π/12 as the exact worked answer, and a decimal as an approximation. A matching decimal at one input is weaker evidence than a derivative identity on the stated domain.

Continue

The written practice asks for both a calculation and a repair of a plausible error. Keep the substitution, bounds and derivative check together. This method belongs to a broader selection process: inspect the numerator, denominator structure and domain before choosing a familiar antiderivative pattern.