Pure · Integration · Year 2

The Reverse Chain Rule

When an integrand contains a function alongside a multiple of its own derivative, the antiderivative can often be written down by inspection — no substitution needed.

9 min read 4 worked examples

Spotting the pattern

The chain rule says $\dfrac{d}{dx}\big[g(f(x))\big] = f'(x)\,g'(f(x))$ — differentiating a composite function always leaves the derivative of the inner function as a factor. So whenever an integrand is built from a function $f(x)$ together with (a multiple of) $f'(x)$, the integral can be reconstructed directly, by recognising which composite function was differentiated to produce it.

The general result

$\displaystyle\int f'(x)\,g'\big(f(x)\big)\,dx = g\big(f(x)\big)+c$

In words: if the integrand is $f'(x)$ multiplied by some function of $f(x)$, the answer is that same "outer" function's antiderivative, evaluated at $f(x)$.

Four common shapes

In practice, four patterns account for almost every reverse-chain-rule integral seen at this level:

Recognise these on sight

$\displaystyle\int f'(x)\,e^{f(x)}\,dx = e^{f(x)}+c \qquad \int \dfrac{f'(x)}{f(x)}\,dx = \ln|f(x)|+c$

$\displaystyle\int f'(x)\big[f(x)\big]^n\,dx = \dfrac{\big[f(x)\big]^{n+1}}{n+1}+c \quad(n\neq-1) \qquad \int f'(x)\cos\big(f(x)\big)\,dx = \sin\big(f(x)\big)+c$

The key skill is identifying $f(x)$ first — usually the "awkward" inner expression — then checking that its derivative $f'(x)$ (up to a constant multiple) really is present elsewhere in the integrand.

Worked example 1

The [f(x)]ⁿ f′(x) pattern

Find $\displaystyle\int 2x(x^2+1)^4\,dx$.

MethodIdentify f(x), confirm f′(x) is present, then write down the antiderivative

Let $f(x)=x^2+1$, so $f'(x)=2x$.

The bracket raised to a power is the natural choice of f(x).

The factor $2x$ outside the bracket is exactly $f'(x)$.

Confirm the derivative is present as a factor, not just a similar-looking term.

$\displaystyle\int 2x(x^2+1)^4\,dx = \dfrac{(x^2+1)^5}{5}+c$.

Apply the [f(x)]ⁿf′(x) result with n = 4.

▪ solved

Worked example 2

The f′(x)eᶠ⁽ˣ⁾ pattern

Find $\displaystyle\int \cos x\,e^{\sin x}\,dx$.

MethodIdentify f(x) inside the exponent

Let $f(x)=\sin x$, so $f'(x)=\cos x$.

The exponent is the natural choice of f(x).

The factor $\cos x$ is exactly $f'(x)$.

Confirm the derivative of the exponent is present as a factor.

$\displaystyle\int \cos x\,e^{\sin x}\,dx = e^{\sin x}+c$.

Apply the f′(x)eᶠ⁽ˣ⁾ result.

▪ solved

Worked example 3

The f′(x)/f(x) pattern

Find $\displaystyle\int \dfrac{6x^2}{x^3+5}\,dx$.

MethodIdentify the denominator as f(x), and match the numerator to a multiple of f′(x)

Let $f(x)=x^3+5$, so $f'(x)=3x^2$.

The denominator is the natural choice of f(x).

The numerator $6x^2 = 2\times3x^2 = 2f'(x)$.

Recognise the numerator as a constant multiple of f′(x).

$\displaystyle\int \dfrac{6x^2}{x^3+5}\,dx = 2\int\dfrac{f'(x)}{f(x)}\,dx = 2\ln|x^3+5|+c$.

Factor out the constant, then apply the f′(x)/f(x) result.

▪ solved

Worked example 4

Recognising f(x) hidden inside a trig power · Challenge

Find $\displaystyle\int \sec^2x\tan^3x\,dx$.

MethodLet f(x) = tan x, and spot its derivative sec²x as the other factor

Let $f(x)=\tan x$, so $f'(x)=\sec^2x$.

tan x appears raised to a power — the natural choice of f(x), even though no bracket is written explicitly.

The factor $\sec^2x$ is exactly $f'(x)$.

Confirm the derivative is present as a factor.

$\displaystyle\int \sec^2x\tan^3x\,dx = \dfrac{\tan^4x}{4}+c$.

Apply the [f(x)]ⁿf′(x) result with n = 3.

▪ solved

f(x) doesn't need brackets

Nothing here is written as an explicit composite function like $(x^2+1)^4$ — yet $\tan x$ raised to a power, with $\sec^2x$ (its own derivative) sitting alongside it, is exactly the same pattern in disguise. Learning to see $f(x)$ inside expressions that aren't obviously "composite" is the real skill this lesson builds.

Common mistake

Applying one of the four results when the "extra" factor is only similar to $f'(x)$, rather than a genuine constant multiple of it. For example, $\displaystyle\int x(x^2+1)^4\,dx$ is not directly of this form, since the outer factor $x$ is only half of $f'(x)=2x$ — spotting that still works here (just factor out $\tfrac12$), but the check has to be made explicitly, not assumed.

Check yourself

Find $\displaystyle\int 3x^2\cos(x^3)\,dx$.

Method

Let $f(x)=x^3$, so $f'(x)=3x^2$.

The angle inside cos is the natural choice of f(x).

The factor $3x^2$ is exactly $f'(x)$.

Confirm the derivative is present as a factor.

$\displaystyle\int 3x^2\cos(x^3)\,dx = \sin(x^3)+c$.

Apply the f′(x)cos(f(x)) result.

▪ solved

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Integrating f(ax + b)

Integrating Using Trigonometric Identities