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ISC 2027
All chaptersMaths · Unit 3

Integrals

6 articles28 formulas32 ways the board asks it
MATStandard Forms & Substitution

Special / Standard Forms

This subtopic is a catalogue of standard integrals built around a quadratic (or its square root) in the denominator: forms reducible to 1a2+x2\dfrac{1}{a^2+x^2}, 1a2−x2\dfrac{1}{a^2-x^2}, 1a2−x2\dfrac{1}{\sqrt{a^2-x^2}}, and their relatives. The universal method is to complete the square in the quadratic, then match a memorised standard result; when a linear numerator is present you split it into (derivative of quadratic) plus a constant.

ISC awards these for clean technique, and they recur inside partial-fraction and definite-integral problems.

Reciprocal of sum / difference of squares
∫dxa2+x2=1atan⁡−1xa+C,∫dxa2−x2=12aln⁡∣a+xa−x∣+C\int \frac{dx}{a^2+x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C, \qquad \int \frac{dx}{a^2-x^2} = \frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right| + C
a>0a>0 constant; the first gives an arctan, the second a logarithm. Complete the square first if the quadratic is not yet in this form.
Reciprocal square-root forms
∫dxa2−x2=sin⁡−1xa+C,∫dxx2±a2=ln⁡∣x+x2±a2∣+C\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a} + C, \qquad \int \frac{dx}{\sqrt{x^2 \pm a^2}} = \ln\left|x+\sqrt{x^2\pm a^2}\right| + C
a>0a>0; the first needs ∣x∣<a|x|<a. For 15−4x−x2\dfrac{1}{\sqrt{5-4x-x^2}} complete the square as 9−(x+2)29-(x+2)^2, an a2−t2a^2-t^2 form with a=3a=3.
Square-root of a quadratic
∫a2−x2 dx=x2a2−x2+a22sin⁡−1xa+C\int \sqrt{a^2-x^2}\,dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C
a>0a>0; companion forms exist for x2±a2\sqrt{x^2\pm a^2}, e.g. ∫9−x2 dx\int\sqrt{9-x^2}\,dx uses a=3a=3.
Linear over quadratic — split the numerator
px+q=p2 (2x+b)+(q−pb2)px+q = \frac{p}{2}\,(2x+b) + \left(q - \frac{pb}{2}\right)
For ∫px+qx2+bx+c dx\int\dfrac{px+q}{x^2+bx+c}\,dx, write the numerator as p2⋅(derivative of denominator)+constant\dfrac{p}{2}\cdot(\text{derivative of denominator}) + \text{constant}, splitting into a ln⁡\ln part and a standard part.
Completing the square
x2+bx+c=(x+b2)2+(c−b24)x^2+bx+c = \left(x+\frac{b}{2}\right)^2 + \left(c - \frac{b^2}{4}\right)
Converts any quadratic into (x+k)2±A2(x+k)^2 \pm A^2 so a standard a2±x2a^2\pm x^2 form applies after the shift t=x+b2t=x+\dfrac{b}{2}.
  • Step 1 for almost every problem here: complete the square in the quadratic to reach (x+k)2±A2(x+k)^2\pm A^2 or A2−(x+k)2A^2-(x+k)^2.
  • For a linear numerator px+qpx+q, set the numerator =λ(denominator’s derivative)+μ= \lambda(\text{denominator's derivative}) + \mu, solve for λ,μ\lambda,\mu, then integrate the two pieces separately.
  • The ln⁡\ln-piece gives ln⁡∣quadratic∣\ln|\text{quadratic}| (or 12ln⁡\dfrac{1}{2}\ln for square-root denominators); the constant-piece gives an arctan/arcsin/log standard form.
  • Distinguish a2+x2a^2+x^2 (arctan) from a2−x2a^2-x^2 (log) and a2−x2\sqrt{a^2-x^2} (arcsin) from x2±a2\sqrt{x^2\pm a^2} (log) — the sign and the square-root decide the answer type.
  • After the shift t=x+b2t=x+\dfrac{b}{2}, the back-substitution must return to xx; here A=∣c−b2/4∣A=\sqrt{|c-b^2/4|} is the effective aa.
  • For ∫dx5−4x−x2\int\dfrac{dx}{5-4x-x^2}, factor out the sign first: 5−4x−x2=9−(x+2)25-4x-x^2 = 9-(x+2)^2, an a2−t2a^2-t^2 form with a=3a=3.
  • Keep a>0a>0 when quoting these results; the formulas assume a positive constant.
Where the marks go
  • Confusing the arctan form 1a2+x2\dfrac{1}{a^2+x^2} with the log form 1a2−x2\dfrac{1}{a^2-x^2} — a sign error flips the entire answer type.
  • Forgetting to factor out a leading negative when completing the square in c−bx−x2c-bx-x^2, leading to a wrong sign under the root.
  • Using sin⁡−1xa\sin^{-1}\dfrac{x}{a} when the radicand is x2+a2x^2+a^2 (which actually gives a ln⁡\ln), or vice-versa.
  • Splitting the linear numerator incorrectly — the coefficient of the derivative term must be p2\dfrac{p}{2} when the derivative is 2x+b2x+b.
How the board asks it
  • Numericalcomplete the square to an arctan/log standard form
    Evaluate ∫dxx2−6x+13\displaystyle\int \dfrac{dx}{x^2 - 6x + 13}.
  • Numericalreciprocal square-root form after the shift (arcsin)
    Evaluate ∫dx5−4x−x2\displaystyle\int \dfrac{dx}{\sqrt{5 - 4x - x^2}}.
  • Numericallinear over quadratic — split numerator into derivative plus constant
    Evaluate ∫2x+3x2+4x+8 dx\displaystyle\int \dfrac{2x + 3}{x^2 + 4x + 8}\,dx.
  • Numericallinear numerator over a square-root of a quadratic
    Evaluate ∫x+2x2+2x+5 dx\displaystyle\int \dfrac{x + 2}{\sqrt{x^2 + 2x + 5}}\,dx.
  • Numericaldefinite integral applying limits to a standard square-root form
    Find ∫01dx3−2x−x2\displaystyle\int_{0}^{1} \dfrac{dx}{\sqrt{3 - 2x - x^2}}.

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.