Integrals of some particular functions

What are the particular integrals?

Use the formulas on intervals within their real domains: the denominators x^² − a^² and a^² − x^² exclude ±a; the square-root denominators require |x| > a and |x| < a, respectively. The plus-sign forms are defined for every real x, with a > 0.

In plain words: these six cover a squared variable and a squared constant, added or subtracted, with or without a square root. Every quadratic that is not a perfect square (b² ≠ 4ac) can be brought to one of these shapes by completing the square, which is why they matter far beyond the six themselves. Two of them give the inverse trigonometric functions tan^⁻¹ and sin^⁻¹ of Chapter 2; the others give logarithms.

Where the formulas come from

NCERT proves all six on pages 244–246. For the substitutions below choose −π/2 < θ < π/2 for tangent or sine, so cos θ > 0. The secant proof uses 0 < θ < π/2 and covers x > a. Its logarithmic answer also holds for x < −a, as direct differentiation verifies; using √(tan^² θ) = tan θ without choosing a branch would be invalid.

• The first two split into simpler fractions: 1/(x² − a²) = (1/(2a))[1/(x − a) − 1/(x + a)], and each part integrates to a logarithm. NCERT’s Note (page 245) says this technique is explained in Section 7.5, on partial fractions.
• tan^⁻¹ and sin^⁻¹: put x = a tan θ in ∫dx/(x² + a²), so that dx = a sec² θ dθ and the integrand becomes (1/a) dθ. In ∫dx/√(a² − x²), put x = a sin θ; the integrand becomes dθ.
• The two logarithms with a root: put x = a sec θ (for √(x² − a²)) or x = a tan θ (for √(x² + a²)). Both lead to ∫sec θ dθ = log |sec θ + tan θ|, which becomes log |x/a + √(x²/a² ∓ 1)|. Taking out −log |a| leaves log |x + √(x² ∓ a²)|, and the constant −log |a| joins C.

A quadratic denominator: complete the square

For ∫dx/(ax² + bx + c) or ∫dx/√(ax² + bx + c), assume the leading coefficient a ≠ 0 and complete the square (NCERT 7.4 (7) and (8), page 246). Write p = b/(2a) and m = c/a − p², that is, m = c/a − b²/(4a²). Then

ax² + bx + c = a[(x + p)² + m].

Now t = x + p gives dx = dt, and the bracket is t² + k² or t² − k², depending on the sign of m. If m = 0, a case NCERT leaves out, it is t² and none of the six is needed: 1/(at²) integrates by the power rule, while under a root, for a > 0, 1/√(at²) is 1/(√a |t|), whose anti derivative is log |t|/√a on t > 0 and −log |t|/√a on t < 0. For a negative leading coefficient under a root, do not take out √a: rewrite the positive quadratic as (−a)(k^² − t^²) and work where |t| < k. Completing the square is the same step as in the Class 11 quadratic formula.

NCERT Examples 8 and 9 (pages 247–249):

• ∫dx/(x² − 16) = (1/8) log |(x − 4)/(x + 4)| + C, with a = 4.
• 2x − x² = 1 − (x − 1)², so ∫dx/√(2x − x²) = sin^⁻¹ (x − 1) + C.
• x² − 6x + 13 = (x − 3)² + 2², so ∫dx/(x² − 6x + 13) = ½ tan^⁻¹ ((x − 3)/2) + C.
• 3x² + 13x − 10 = 3[(x + 13/6)² − (17/6)²], and the first formula gives (1/17) times log |(3x − 2)/(x + 5)|, plus C. NCERT’s last step drops the constant (1/17) log (1/3) into C.

A linear numerator

For ∫(px + q)/(ax² + bx + c) dx, and the same with a square root below, write the top as a multiple of the derivative of the bottom plus a constant (NCERT 7.4 (9) and (10), pages 246–247):

Equate the coefficients of x and the constants to find A and B. The A part has the derivative of the bottom on top, so it gives a logarithm (or a square root, when the bottom has a root); the B part is a quadratic denominator of the previous section.

NCERT Example 10 (i) (pages 249–250): x + 2 = A(4x + 6) + B gives A = ¼ and B = ½, and

∫(x + 2)/(2x² + 6x + 5) dx

= ¼ log |2x² + 6x + 5| + ½ tan^⁻¹ (2x + 3) + C.

NCERT Example 10 (ii) (pages 250–251): x + 3 = A(−4 − 2x) + B gives A = −½ and B = 1. With a root below, the A part is a square root, not a logarithm, and 5 − 4x − x² = 9 − (x + 2)²:

∫(x + 3)/√(5 − 4x − x²) dx

= −√(5 − 4x − x²) + sin^⁻¹ ((x + 2)/3) + C.

Solved examples

Common mistakes to avoid

Quick revision summary

• ∫dx/(x^² + a^²) = (1/a) tan^⁻¹ (x/a) + C and ∫dx/√(a^² − x^²) = sin^⁻¹ (x/a) + C; the other four of the six give logarithms.
• A quadratic below: take out a, complete the square, put t = x + b/(2a), and use the matching formula.
• A linear numerator: px + q = A(2ax + b) + B; the A part gives a logarithm (or a root), the B part a quadratic denominator.

Questions students ask

Why is there a 1/a in the tan^⁻¹ formula but not in the sin^⁻¹ formula?

∫dx/(x² + a²) = (1/a) tan^⁻¹ (x/a) + C has a 1/a but ∫dx/√(a² − x²) = sin^⁻¹ (x/a) + C has none because of how a scales out (a > 0): putting x = au makes dx = a du, while x² + a² = a²(u² + 1) carries two factors of a and √(a² − x²) = a√(1 − u²) only one. So one a stays below in the first, (1/a)∫du/(u² + 1), and none in the second, ∫du/√(1 − u²). For a = 5, ∫dx/(x² + 25) is (1/5) tan^⁻¹ (x/5) + C and ∫dx/√(25 − x²) is sin^⁻¹ (x/5) + C.

How does the discriminant decide which formula ∫dx/(ax² + bx + c) needs?

The sign of the discriminant D = b² − 4ac decides it, because completing the square writes ax² + bx + c as a(X² − D/(4a²)), with X = x + b/(2a): a positive D leaves a difference of squares and a logarithm, a negative D a sum of squares and tan^⁻¹. A zero D gives a perfect square, which none of the six formulas covers; the power rule does it: ∫dx/(x² − 8x + 16) is ∫(x − 4)⁻² dx, that is −1/(x − 4) + C. For x² + 4x + 7, D = −12 and the square is (x + 2)² + 3, so the integral is (1/√3) tan^⁻¹ w + C, where w = (x + 2)/√3.

What is the integral of 1/√(x² + a²), and is it the same as sinh^⁻¹ (x/a)?

The integral of 1/√(x² + a²) is log |x + √(x² + a²)| plus C (NCERT page 244); beyond the NCERT text, which has no hyperbolic functions, sinh^⁻¹ (x/a) + C, the inverse hyperbolic sine that many calculus books and computer algebra tools give, is the same answer up to a constant. Since sinh^⁻¹ u equals log (u + √(u² + 1)), with u = x/a and a > 0 it is log (x + √(x² + a²)) minus log a, and the − log a joins C. In the same way cosh^⁻¹ (x/a) matches log |x + √(x² − a²)| for x ≥ a; NCERT writes the logarithms.

When does ∫dx/√(ax² + bx + c) give sin^⁻¹ and when a logarithm?

For ∫dx/√(ax² + bx + c) the sign of a decides: completing the square with X = x + b/(2a), a > 0 gives a(X² ± k²) and a logarithm, while a < 0 gives −a(k² − X²) and sin^⁻¹. For a < 0 the discriminant b² − 4ac must also be positive, or the quadratic is never positive, as with −x² − 1. So ∫dx/√(10x − x² − 21), where 10x − x² − 21 equals 4 − (x − 5)², is sin^⁻¹ ((x − 5)/2) + C, while ∫dx/√(x² − 4x + 13), where the square is (x − 2)² + 9, is log |x − 2 + √(x² − 4x + 13)| + C.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Find ∫dx/√(1 + 4x²) and ∫dx/√(9 − 25x²) (NCERT Exercise 7.4, Q2 and Q4, page 251).

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   With t = 2x: ½∫dt/√(1 + t²) = ½ log |2x + √(1 + 4x²)| + C. With t = 5x: (1/5)∫dt/√(9 − t²) = (1/5) sin^⁻¹ (5x/3) + C.

2. Q2. Find ∫dx/√(x² + 2x + 2) (NCERT Exercise 7.4, Q10, page 252).

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   x² + 2x + 2 = (x + 1)² + 1, so the integral is log |x + 1 + √(x² + 2x + 2)| + C.

3. Q3. Find ∫dx/√(7 − 6x − x²) (NCERT Exercise 7.4, Q12, page 252).

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   7 − 6x − x² = 16 − (x + 3)², so the integral is sin^⁻¹ ((x + 3)/4) + C.

4. Q4. ∫dx/(x² + 2x + 2) equals (A) x tan^⁻¹ (x + 1) + C; (B) tan^⁻¹ (x + 1) + C; (C) (x + 1) tan^⁻¹ x + C; (D) tan^⁻¹ x + C (NCERT Exercise 7.4, Q24, page 252).

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   x² + 2x + 2 = (x + 1)² + 1, so the integral is tan^⁻¹ (x + 1) + C: (B).