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Square-root integrals look alike and are easy to mix up: the one with a sin⁻¹, the one with a log and a plus sign, the one with a log and a minus sign. This post gives the three results, shows where one of them comes from so you can rebuild it, and works three examples where the real skill is completing the square into the right form. Every answer here was checked by differentiating it back.
Are these integrals in the syllabus?
Yes, by name. The JEE (Main) 2026 syllabus lists "simple integrals of the type" ending with ∫√(a² ± x²) dx and ∫√(x² − a²) dx in Unit 8. NCERT's Class 12 chapter Integrals (2026-27 reprint) derives them in Section 7.6.2, "Integrals of some more types", with Examples 23 and 24 and Exercise 7.7. The JEE (Advanced) 2026 brochure covers integration by parts and by substitution, which is all these need.
What are the three results?
∫√(x² − a²) dx = (x/2)√(x² − a²) − (a²/2) log|x + √(x² − a²)| + C
∫√(x² + a²) dx = (x/2)√(x² + a²) + (a²/2) log|x + √(x² + a²)| + C
∫√(a² − x²) dx = (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) + C
A way to hold them: the first term is always (x/2) times the root. The second term has a²/2 in front. If a² is added inside the root, the log term is added; if a² is subtracted, the log term is subtracted. If x² carries the minus sign, the log becomes sin⁻¹(x/a).
Where does the sin⁻¹ version come from?
NCERT takes the constant function 1 as the second function and integrates by parts. For I = ∫√(a² − x²) dx:
I = x√(a² − x²) − ∫x · (−x)/√(a² − x²) dx = x√(a² − x²) + ∫x²/√(a² − x²) dx.
Write x² = a² − (a² − x²): the integral becomes ∫a²/√(a² − x²) dx − ∫√(a² − x²) dx = a² sin⁻¹(x/a) − I.
So 2I = x√(a² − x²) + a² sin⁻¹(x/a), which gives the result.
Geometry gives a free check. The curve y = √(a² − x²) is the upper half of a circle of radius a, so the integral from 0 to a should be a quarter circle, πa²/4. The formula gives (a/2)(0) + (a²/2)(π/2) − 0 = πa²/4.
Worked example 1: ∫√(8 + 2x − x²) dx
Complete the square: 8 + 2x − x² = 9 − (x² − 2x + 1) = 9 − (x − 1)².
So a = 3 and the variable is x − 1. The x² term is negative, so this is the sin⁻¹ form.
Answer: ((x − 1)/2)√(8 + 2x − x²) + (9/2) sin⁻¹((x − 1)/3) + C.
Area check: y = √(9 − (x − 1)²) is a semicircle of radius 3 centred at x = 1, from x = −2 to x = 4. Its area is (1/2)π(3²) = 9π/2. Substituting the limits into the answer gives (9/2)(π/2) − (9/2)(−π/2) = 9π/2.
Worked example 2: ∫√(x² − 6x + 13) dx
Complete the square: x² − 6x + 13 = (x − 3)² + 4.
So a = 2 and the form is √(y² + a²) with y = x − 3.
Answer: ((x − 3)/2)√(x² − 6x + 13) + 2 log|x − 3 + √(x² − 6x + 13)| + C. The 2 in front of the log is a²/2 = 4/2.
Worked example 3: a coefficient on x²
∫√(4x² + 9) dx. The formulas need x² with coefficient 1, so take the 4 out first:
√(4x² + 9) = 2√(x² + 9/4), so a = 3/2.
∫ = 2[(x/2)√(x² + 9/4) + (9/8) log|x + √(x² + 9/4)|] + C.
Simplify: (x/2)√(4x² + 9) + (9/4) log|2x + √(4x² + 9)| + C. Moving the factor 2 inside the log only changes the constant.
Which slips cost marks?
Forgetting to take out the coefficient of x², as in example 3.
Writing sin⁻¹(x/a²) or sin⁻¹(x²/a) instead of sin⁻¹(x/a).
Getting the sign of the log term wrong for √(x² − a²).
Completing the square with a negative leading term carelessly: 8 + 2x − x² is 9 − (x − 1)², not (x − 1)² − 9.
Dropping the 1/2 in front of each term.
When you get a different-looking answer from the key, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can see whether you differ by a constant or by a real error.
A practice routine
Derive the √(x² + a²) result by parts yourself, following the three steps above.
Do NCERT Exercise 7.7 questions 1 to 9 without looking at the formulas, then check.
For each answer, differentiate it once to see the integrand come back.
Do one definite integral where geometry gives the answer, such as ∫ from 0 to 2 of √(4 − x²) dx = π.
If quadratic denominators still trip you up, our post on integrals with quadratic denominators uses the same completing-the-square step. Start tonight by writing the three results from memory and checking them against this page.
