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L = ∫_a^b √1 + [dy/dx]² dx, or ∫ √1 + [dx/dy]² dy when the curve is written x = g(y).

Parametric: L = ∫ √[x′(t)]² + [y′(t)]² dt.

Polar: L = ∫ √r² + [dr/dθ]² dθ.

The perfect-square trick: if y′ = a·x^p − 14ax^−p then the cross terms cancel to +½, so 1 + [y′]² = (a·x^p + 14ax^−p)² and the radical disappears.

Identities that clear the radical: 1 + tan²θ = sec²θ · 1 + cot²θ = csc²θ · 1 + sinh²u = cosh²u.

Worth memorising: ∫sec θ dθ = ln|sec θ + tan θ| · ∫√1+u² du = ½(u√1+u² + ln|u + √1+u²|).

Answers: type an exact value like 17/12, 3/4 + ln(2)/4 or sqrt(2)+ln(1+sqrt(2)), or a decimal.

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