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Pure · Integration

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Integration by reversing the chain rule

Spotting integrals of the form ∫f'(x)[f(x)]ⁿ, ∫f'(x)/f(x) → ln|f(x)| and ∫f'(x)e^{f(x)} — "integration by recognition", where the derivative of the inside appears as a factor.

Pure · Integration

Integration by reversing the chain rule

Spotting integrals of the form ∫f'(x)[f(x)]ⁿ, ∫f'(x)/f(x) → ln|f(x)| and ∫f'(x)e^{f(x)} — "integration by recognition", where the derivative of the inside appears as a factor.

Why it works

The chain rule says ddx g(f(x))=g′(f(x)) f′(x)\dfrac{d}{dx}\,g(f(x)) = g'(f(x))\,f'(x) — the derivative of the outer function times the derivative of the inside. Run that backwards and a powerful pattern appears: if the derivative of the inside function is sitting in the integrand as a factor, you can integrate by recognition.

The three patterns worth knowing on sight:
  • A power of a function: ∫f′(x) [f(x)]n dx=[f(x)]n+1n+1+c\displaystyle\int f'(x)\,[f(x)]^n\,dx = \frac{[f(x)]^{n+1}}{n+1} + c
(n≠−1)(n \ne -1). For example ∫2x(x2+1)3 dx=(x2+1)44+c\int 2x(x^2 + 1)^3\,dx = \dfrac{(x^2+1)^4}{4} + c, because 2x2x is exactly the derivative of the inside x2+1x^2 + 1.
  • The n=−1n = -1 case — a log: ∫f′(x)f(x) dx=ln⁡∣f(x)∣+c\displaystyle\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + c.
When the top is the derivative of the bottom, the integral is the log of the bottom. For example ∫2xx2+1 dx=ln⁡(x2+1)+c\int \dfrac{2x}{x^2 + 1}\,dx = \ln(x^2 + 1) + c.
  • An exponential of a function: ∫f′(x) ef(x) dx=ef(x)+c\displaystyle\int f'(x)\,e^{f(x)}\,dx = e^{f(x)} + c.
For example ∫cos⁡x esin⁡x dx=esin⁡x+c\int \cos x\,e^{\sin x}\,dx = e^{\sin x} + c.

Adjusting the constant. Often the derivative of the inside is there *up to a number*. Then write down the natural answer and fix the constant by differentiating back. For ∫x(x2+1)3 dx\int x(x^2 + 1)^3\,dx the inside needs 2x2x but you only have xx — so try 14(x2+1)4\tfrac14(x^2+1)^4, differentiate to get 2x(x2+1)32x(x^2+1)^3, see it is twice too big, and halve it: 18(x2+1)4+c\tfrac18(x^2+1)^4 + c. This "guess and adjust" is the heart of recognition.

If the derivative of the inside is not a constant multiple of what is present — say ∫x2(x2+1)3 dx\int x^2(x^2+1)^3\,dx, where you'd need 2x2x but have x2x^2 — recognition fails and you need a different method (substitution, or expanding).