9 min readConcept sheet1-mark MCQ and 2-mark questions; these six results sit inside most 3-mark and 5-mark integrals

Special integrals: 1/(x² ± a²), 1/(a² − x²) and their square roots

Six results handle sums and differences of squares. Two give tan⁻¹ or sin⁻¹; the other four give logs. Learn them as one set.

tan⁻¹ form∫ dx/(x² + a²) = (1/a) tan⁻¹(x/a) + C

Log forms without a root∫ dx/(x² − a²) = (1/2a) log|(x − a)/(x + a)| + C; ∫ dx/(a² − x²) = (1/2a) log|(a + x)/(a − x)| + C

Forms with a root∫ dx/√(a² − x²) = sin⁻¹(x/a) + C; ∫ dx/√(x² ± a²) = log|x + √(x² ± a²)| + C

Symbolsa is a positive constant, and (1/2a) means 1/(2a)

The six results

Integrand	Integral	Memory hook
1/(x² − a²)	(1/2a) log\|(x − a)/(x + a)\| + C	x first, so (x − a) goes on top
1/(a² − x²)	(1/2a) log\|(a + x)/(a − x)\| + C	a first, so (a + x) goes on top
1/(x² + a²)	(1/a) tan⁻¹(x/a) + C	A sum of squares gives tan⁻¹
1/√(x² − a²)	log\|x + √(x² − a²)\| + C	x plus the same root
1/√(x² + a²)	log\|x + √(x² + a²)\| + C	x plus the same root
1/√(a² − x²)	sin⁻¹(x/a) + C	A root of (a² − x²) gives sin⁻¹

Proving two of them

Your textbook proves all six. These two show the two main ideas: partial fractions and a trigonometric substitution.

1
1/(x² − a²) = 1/((x − a)(x + a)) = (1/2a)[1/(x − a) − 1/(x + a)].
2
So ∫ dx/(x² − a²) = (1/2a)[log|x − a| − log|x + a|] + C = (1/2a) log|(x − a)/(x + a)| + C.
3
For ∫ dx/(x² + a²), put x = a tan θ. Then dx = a sec² θ dθ.
4
x² + a² = a² tan² θ + a² = a² sec² θ.
5
The integral becomes ∫ (a sec² θ)/(a² sec² θ) dθ = (1/a) ∫ dθ = θ/a + C.
6
Since θ = tan⁻¹(x/a): so, ∫ dx/(x² + a²) = (1/a) tan⁻¹(x/a) + C.

Making an integrand fit

If x² has a coefficient, take it out first: 9 − 4x² = 4[(3/2)² − x²]. Or put t = 2x, so dx = dt/2.
If x⁴ or x⁶ appears with x or x² as a factor, put t = x² or t = x³ first.
A sum of squares with no root gives tan⁻¹. A root of (a² − x²) gives sin⁻¹. All the others give log.
Do not mix up 1/(x² + a²) (a tan⁻¹ result) with 1/√(x² + a²) (a log result).

Example 1: a root with x² + a²

Problem

Find ∫ dx/√(4x² + 25).

Working

Put t = 2x. Then dt = 2 dx, so dx = dt/2.
4x² + 25 = t² + 5².
The integral becomes (1/2) ∫ dt/√(t² + 5²).
Use ∫ dt/√(t² + a²) = log|t + √(t² + a²)| + C with a = 5.
= (1/2) log|t + √(t² + 25)| + C.
Put back t = 2x.

Answer

(1/2) log|2x + √(4x² + 25)| + C

Example 2: a² − x² with no root

Problem

Find ∫ dx/(16 − 9x²).

Working

Put t = 3x. Then dx = dt/3, and 16 − 9x² = 4² − t².
The integral becomes (1/3) ∫ dt/(4² − t²).
Use ∫ dt/(a² − t²) = (1/2a) log|(a + t)/(a − t)| + C with a = 4.
= (1/3)(1/8) log|(4 + t)/(4 − t)| + C.
Put back t = 3x.

Answer

(1/24) log|(4 + 3x)/(4 − 3x)| + C

Example 3: substitute first (board style)

Problem

Find ∫ x²/(x⁶ + 16) dx.

Working

Notice x⁶ = (x³)², and x² is a constant times the derivative of x³.
Put t = x³. Then dt = 3x² dx, so x² dx = dt/3.
The integral becomes (1/3) ∫ dt/(t² + 4²).
Use ∫ dt/(t² + a²) = (1/a) tan⁻¹(t/a) + C with a = 4.
= (1/3)(1/4) tan⁻¹(t/4) + C.
Put back t = x³.

Answer

(1/12) tan⁻¹(x³/4) + C

In real life

An electric line hanging between two pylons sags into a curve called a catenary. Finding its shape uses ∫ dp/√(1 + p²) = log|p + √(1 + p²)|, one of these six results.

Exam angle

1 marks: “∫ dx/(x² + 16) equals:” Use a = 4: (1/4) tan⁻¹(x/4) + C.

2 marks: “Find ∫ eˣ/√(1 − e²ˣ) dx.” Put t = eˣ, dt = eˣ dx. ∫ dt/√(1 − t²) = sin⁻¹ t + C. Answer: sin⁻¹(eˣ) + C.

2 marks: “Find ∫ 3x²/(x⁶ − 1) dx.” Put t = x³, dt = 3x² dx. ∫ dt/(t² − 1) = (1/2) log|(t − 1)/(t + 1)| + C. Answer: (1/2) log|(x³ − 1)/(x³ + 1)| + C.

Remember it

Sum of squares: tan⁻¹. Root of a² minus x²: sin⁻¹. Everything else: log.