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CBSE · Class 12 · Mathematics

Integrals of Some Particular Functions (Ch 7: 7.4)

3 concepts · 24 practice questions with answers · Free to read

Six standard forms

From daily life

A mason keeps a few ready-made moulds for common shapes. In the same way, six integrals of the type 1/(x² ± a²) and 1/√(a² ± x²) are used as moulds for many harder integrals.

Simple definition

These integrals involve a sum or difference of squares in the denominator. Learn the six results, then match the integrand with the right one by finding a.

Formula: ∫ dx/(x² + a²) = (1/a) tan⁻¹(x/a) + C, ∫ dx/(x² − a²) = (1/2a) ln|(x − a)/(x + a)| + C, ∫ dx/(a² − x²) = (1/2a) ln|(a + x)/(a − x)| + C

Rule 1: Squares in the denominator — Find a, then pick the sign.

  • Easy: ∫ dx/(x² + 9) = (1/3) tan⁻¹(x/3) + C.
  • Medium: ∫ dx/(x² − 16) = (1/8) ln|(x − 4)/(x + 4)| + C.
  • Exam-type: ∫ dx/(4 − x²) = (1/4) ln|(2 + x)/(2 − x)| + C.

Rule 2: Square root in the denominator — ∫ dx/√(a² − x²) = sin⁻¹(x/a) + C, ∫ dx/√(x² + a²) = ln|x + √(x² + a²)| + C, ∫ dx/√(x² − a²) = ln|x + √(x² − a²)| + C.

  • Easy: ∫ dx/√(25 − x²) = sin⁻¹(x/5) + C.
  • Medium: ∫ dx/√(x² + 4) = ln|x + √(x² + 4)| + C.
  • Exam-type: ∫ dx/√(x² − 9) = ln|x + √(x² − 9)| + C.
Remember

In x² + a² the a is the square root of the constant, so for x² + 25 take a = 5.

Common mistake

Writing ∫ dx/(x² + 9) = (1/9) tan⁻¹(x/9) instead of (1/3) tan⁻¹(x/3).

Memory clue

Plus gives tan⁻¹, minus gives ln, root with minus gives sin⁻¹.

Practice questions

Easy
  1. 1Find ∫ dx/(x² + 25).
  2. 2Find ∫ dx/(x² − 4).
  3. 3Find ∫ dx/√(9 − x²).
Medium
  1. 4Find ∫ dx/(9 − x²).
  2. 5Find ∫ dx/√(x² + 16).
  3. 6Find ∫ dx/√(x² − 1).
Challenging
  1. 7Find ∫ dx/(1 + 4x²).
  2. 8Find ∫ dx/√(4 − 9x²).
Show answers (try them yourself first)
  1. 1(1/5) tan⁻¹(x/5) + C
  2. 2(1/4) ln|(x − 2)/(x + 2)| + C
  3. 3sin⁻¹(x/3) + C
  4. 4(1/6) ln|(3 + x)/(3 − x)| + C
  5. 5ln|x + √(x² + 16)| + C
  6. 6ln|x + √(x² − 1)| + C
  7. 7(1/2) tan⁻¹(2x) + C
  8. 8(1/3) sin⁻¹(3x/2) + C

Completing the square

From daily life

Folding a crooked sheet along its middle makes both halves match. Completing the square folds a quadratic into the shape (x + p)² ± a², and then a standard form applies.

Simple definition

When the denominator contains ax² + bx + c, write it as a perfect square plus or minus a constant, put t = x + p, and use one of the six forms. If the numerator is px + q, split it into a multiple of the derivative of the quadratic plus a constant.

Formula: x² + 2px + c = (x + p)² + (c − p²)

Rule 1: Complete the square, then match — Shift x by p.

  • Easy: ∫ dx/(x² + 4x + 13) = ∫ dx/((x + 2)² + 9) = (1/3) tan⁻¹((x + 2)/3) + C.
  • Medium: ∫ dx/(x² − 6x + 5) = ∫ dx/((x − 3)² − 4) = (1/4) ln|(x − 5)/(x − 1)| + C.
  • Exam-type: ∫ dx/√(5 − 4x − x²) = ∫ dx/√(9 − (x + 2)²) = sin⁻¹((x + 2)/3) + C.

Rule 2: Numerator px + q — Write it as (A × derivative of the quadratic) + B.

  • Easy: ∫ (2x + 3)/(x² + 3x + 5) dx = ln(x² + 3x + 5) + C.
  • Medium: ∫ (x + 1)/(x² + 4x + 5) dx = ½ ln(x² + 4x + 5) − tan⁻¹(x + 2) + C.
  • Exam-type: ∫ dx/(4x² + 4x + 5) = ∫ dx/((2x + 1)² + 4) = ¼ tan⁻¹((2x + 1)/2) + C.
Remember

Take out the coefficient of x² first. For 4x² + 4x + 5 the factor 2 from 2x + 1 appears in the final denominator.

Common mistake

Forgetting to divide by the coefficient of x when t = 2x + 1 is substituted.

Memory clue

Square it up, shift x, then use the six forms.

Practice questions

Easy
  1. 1Find ∫ dx/(x² + 4x + 13).
  2. 2Find ∫ dx/(x² − 6x + 5).
  3. 3Find ∫ dx/(x² + 2x + 2).
Medium
  1. 4Find ∫ dx/√(5 − 4x − x²).
  2. 5Find ∫ dx/√(x² + 2x + 5).
  3. 6Find ∫ (2x + 3)/(x² + 3x + 5) dx.
Challenging
  1. 7Find ∫ (x + 1)/(x² + 4x + 5) dx.
  2. 8Find ∫ dx/(4x² + 4x + 5).
Show answers (try them yourself first)
  1. 1(1/3) tan⁻¹((x + 2)/3) + C
  2. 2(1/4) ln|(x − 5)/(x − 1)| + C
  3. 3tan⁻¹(x + 1) + C
  4. 4sin⁻¹((x + 2)/3) + C
  5. 5ln|x + 1 + √(x² + 2x + 5)| + C
  6. 6ln(x² + 3x + 5) + C
  7. 7½ ln(x² + 4x + 5) − tan⁻¹(x + 2) + C
  8. 8¼ tan⁻¹((2x + 1)/2) + C

Square-root forms

From daily life

The area of a half-disc cannot be found by adding simple slices of a straight line. The integrals of √(a² − x²) and its relatives give such curved areas exactly.

Simple definition

Integrals of the types √(a² − x²), √(x² + a²) and √(x² − a²) are worked out by parts. Remember their final results and use them directly.

Formula: ∫ √(a² − x²) dx = (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) + C

Rule 1: Three results — Pick the one with the matching signs.

  • Easy: ∫ √(16 − x²) dx = (x/2)√(16 − x²) + 8 sin⁻¹(x/4) + C.
  • Medium: ∫ √(x² + 9) dx = (x/2)√(x² + 9) + (9/2) ln|x + √(x² + 9)| + C.
  • Exam-type: ∫ √(x² − 4) dx = (x/2)√(x² − 4) − 2 ln|x + √(x² − 4)| + C.

Rule 2: Limits — A definite integral of √(a² − x²) gives the area of part of a circle.

  • Easy: ∫ from −3 to 3 of √(9 − x²) dx = 9π/2, the area of a half disc of radius 3.
  • Medium: ∫ from 0 to 2 of √(4 − x²) dx = π, a quarter disc of radius 2.
  • Exam-type: ∫ from 0 to 1 of √(1 − x²) dx = π/4.
Remember

In the last term, the coefficient is a²/2 and the inverse sine has x/a inside.

Common mistake

Using a instead of a² in the coefficient of the second term.

Memory clue

x over two times the root, plus a-square over two times the inverse function.

Practice questions

Easy
  1. 1Find ∫ √(16 − x²) dx.
  2. 2Find ∫ √(x² + 9) dx.
  3. 3Find ∫ √(x² − 4) dx.
Medium
  1. 4Find ∫ √(1 − x²) dx.
  2. 5Find ∫ √(25 − x²) dx.
  3. 6Find ∫ √(x² + 1) dx.
Challenging
  1. 7Find the value of ∫ from −3 to 3 of √(9 − x²) dx.
  2. 8Find the value of ∫ from 0 to 2 of √(4 − x²) dx.
Show answers (try them yourself first)
  1. 1(x/2)√(16 − x²) + 8 sin⁻¹(x/4) + C
  2. 2(x/2)√(x² + 9) + (9/2) ln|x + √(x² + 9)| + C
  3. 3(x/2)√(x² − 4) − 2 ln|x + √(x² − 4)| + C
  4. 4(x/2)√(1 − x²) + ½ sin⁻¹x + C
  5. 5(x/2)√(25 − x²) + (25/2) sin⁻¹(x/5) + C
  6. 6(x/2)√(x² + 1) + ½ ln|x + √(x² + 1)| + C
  7. 79π/2 (area of a half disc of radius 3)
  8. 8π (area of a quarter disc of radius 2)