Lesson 6 of 13 · 7 min
Integrals of some particular functions
NCERT §7.4
Denominators such as x² + 4, 9 − x² and x² − 6x + 13 appear again and again. Each is a square plus or minus a constant, and six formulas cover them all.
The lesson in notes
In short
Six results, all checkable by differentiation, handle a squared term plus or minus a constant a² (a > 0).
∫dx/(x² − a²) = (1/2a) log|(x − a)/(x + a)| + C and ∫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, and ∫dx/√(x² + a²) = log|x + √(x² + a²)| + C.
A general quadratic ax² + bx + c in the denominator, with or without a square root, is first completed to a square: it becomes a[(x + b/2a)² ± k²], and one of the six results applies.
Worked example: x² − 6x + 13 = (x − 3)² + 2², so ∫dx/(x² − 6x + 13) = ½ tan⁻¹((x − 3)/2) + C.
A linear numerator over a quadratic: write px + q = A × (derivative of the quadratic) + B. The A part integrates to a logarithm (or a square root, under a root), and the B part is the previous type.
Worked example: x + 2 = ¼(4x + 6) + ½, so ∫(x + 2)/(2x² + 6x + 5) dx = ¼ log|2x² + 6x + 5| + ½ tan⁻¹(2x + 3) + C.