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Integrals like ∫ dx/(x² + 4x + 13) look different every time, but they all reduce to six standard results with one technique: completing the square. The step that goes wrong is usually not the integration but the algebra before it, a dropped coefficient or a wrong sign. This post shows the decision rule, four worked examples and the checks to run.
Which standard forms do you need?
The JEE (Main) 2026 syllabus lists integration by substitution and "evaluation of simple integrals" in Unit 8. Section 7.4 of the current NCERT Class 12 chapter on integrals lists the six results:
∫ dx/(x² − a²) = (1/2a) log|(x − a)/(x + a)| + C
∫ dx/(a² − x²) = (1/2a) log|(a + x)/(a − x)| + C
∫ dx/(x² + a²) = (1/a) tan⁻¹(x/a) + C
∫ dx/√(x² − a²) = log|x + √(x² − a²)| + C
∫ dx/√(a² − x²) = sin⁻¹(x/a) + C
∫ dx/√(x² + a²) = log|x + √(x² + a²)| + C
NCERT then shows that ax² + bx + c = a[(x + b/2a)² + (c/a − b²/4a²)], so the substitution t = x + b/2a turns any quadratic denominator into one of these.
How do you know which form you will get?
Look at the discriminant, b² − 4ac, before you start.
Negative: the quadratic has no real roots, and completing the square gives a sum of squares. Expect tan⁻¹ (or the log with a plus sign under a root).
Positive: it factorises, and you get a difference of squares. Expect a log. Partial fractions would also work.
Zero: it is a perfect square. ∫ dx/(x + h)² = −1/(x + h) + C, with no standard form needed.
Worked example 1: a sum of squares
Find ∫ dx/(x² + 4x + 13). The discriminant is 16 − 52 < 0, so expect tan⁻¹.
x² + 4x + 13 = (x + 2)² + 9. With t = x + 2 and a = 3: (1/3) tan⁻¹((x + 2)/3) + C.
Worked example 2: a difference of squares
Find ∫ dx/(x² − 6x + 5). The discriminant is 36 − 20 > 0, so expect a log.
x² − 6x + 5 = (x − 3)² − 4. With t = x − 3 and a = 2: (1/4) log|(x − 3 − 2)/(x − 3 + 2)| + C = (1/4) log|(x − 5)/(x − 1)| + C.
Worked example 3: a linear numerator
Find ∫ (2x + 3)/(x² + 2x + 5) dx.
NCERT's method: write px + q = A × (derivative of the denominator) + B. Here 2x + 3 = A(2x + 2) + B. Comparing coefficients, A = 1 and B = 1.
So the integral splits into ∫ (2x + 2)/(x² + 2x + 5) dx + ∫ dx/(x² + 2x + 5).
The first part is log(x² + 2x + 5), with no modulus needed because the quadratic is always positive. For the second, x² + 2x + 5 = (x + 1)² + 4, giving (1/2) tan⁻¹((x + 1)/2).
Answer: log(x² + 2x + 5) + (1/2) tan⁻¹((x + 1)/2) + C.
Worked example 4: under a square root
Find ∫ dx/√(6x − x² − 5).
Here a = −1, so take the minus sign out carefully: 6x − x² − 5 = −(x² − 6x + 5) = −[(x − 3)² − 4] = 4 − (x − 3)².
That is a² − t² with a = 2, so the answer is sin⁻¹((x − 3)/2) + C.
Differentiate to check: the derivative of sin⁻¹((x − 3)/2) is (1/2)/√(1 − (x − 3)²/4) = 1/√(4 − (x − 3)²). Correct.
Which mistakes cost the most marks?
Completing the square with the wrong constant: x² + 4x + 13 is (x + 2)² + 9, not (x + 2)² + 13.
Forgetting to take out a negative leading coefficient, which flips a sum of squares into a difference.
Mixing up the two log forms. Check by differentiating, or by noting the log must blow up where the denominator is zero.
Splitting the numerator with the derivative of the wrong quadratic.
Dropping the 1/a in front of tan⁻¹.
A practice routine
Discriminant first (10 minutes, three times a week). For ten quadratic denominators, write only the sign of b² − 4ac and the form you expect.
Complete the square on its own. Five quadratics a day, including ones with a ≠ 1 and a < 0.
Split numerators. Three integrals a week of the (px + q) type.
Mix it in with partial fractions and substitution, so you practise choosing. In Rohrer and Taylor's experiment, mixed practice gave "vastly superior" results a week later.
If you forget why one form gives tan⁻¹ and another a log, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, each with its own lesson.
Today, redo worked example 3 with numerator 3x − 1 and check your answer by differentiating.
