P3 5.2 - Recognising reverse chain rules | Edexcel A-Level International Mathematics

Recognise integrands built from an inner function and its derivative. This lets you integrate composite powers, logarithmic forms and trigonometric expressions by undoing differentiation, sometimes after an identity makes the pattern visible.

Recognising f-prime times a power of f

Can $\int6x(x^2+1)^2\,dx$ be found without expanding the brackets? Differentiating a power of $x^2+1$ produces both a power of that bracket and a factor of $x$ . That is the pattern to reverse.

The chain rule says

\frac{d}{dx}[f(x)]^{n+1}=(n+1)[f(x)]^n f'(x).

So look for the inner function $f(x)$ , its power, and a multiplier proportional to $f'(x)$ . You can supply a missing constant factor and compensate outside the integral; you cannot supply a missing factor of $x$ without changing the problem.

Reverse chain rule for powers

For $n\ne-1$ , on an interval where the expressions are defined,

\int f'(x)[f(x)]^n\,dx=\frac{[f(x)]^{n+1}}{n+1}+c.

Multiply the result by any constant multiplier present in the integrand.

Not in the formula booklet: learn it.

Predict the new bracket power first. Then determine the coefficient by differentiating it.

Worked example 1: a constant multiple of the derivative

Find $\int6x(x^2+1)^2\,dx$ .

Plan. Identify $f(x)=x^2+1$ and write $6x=3f'(x)=3(2x)$ .

\begin{aligned} \int6x(x^2+1)^2\,dx &=3\int2x(x^2+1)^2\,dx\\ &=3\left(\frac{(x^2+1)^3}{3}\right)+c \end{aligned}

Why these factors? The outer power rises from $2$ to $3$ , so divide by $3$ ; the integrand already contains three times the inner derivative, so multiply by $3$ .

Answer. $(x^2+1)^3+c$ . This answers the opening question without expansion.

Check. Its derivative is $3(x^2+1)^2(2x)=6x(x^2+1)^2$ .

Change only the bracket power from $2$ to $3/2$ . The new power is $5/2$ , so the numerical compensation changes too.

Worked example 2: a fractional bracket power

Find $\int6x(x^2+1)^{3/2}\,dx$ .

Plan. Keep the factor $3(2x)$ , raise the bracket power by one, and divide by the new power $5/2$ .

\begin{aligned} \int6x(x^2+1)^{3/2}\,dx &=3\int2x(x^2+1)^{3/2}\,dx\\ &=3\left(\frac{(x^2+1)^{5/2}}{5/2}\right)+c\\ &=\frac65(x^2+1)^{5/2}+c \end{aligned}

The bracket $x^2+1$ is positive for every real $x$ , so this fractional power introduces no further domain restriction.

Answer. $\frac65(x^2+1)^{5/2}+c$ .

Check. Differentiating gives $\frac65\times\frac52(x^2+1)^{3/2}\times2x=6x(x^2+1)^{3/2}$ .

Negative powers also fit when $n\ne-1$ . For example, raising $-2$ by one gives $-1$ , with division by $-1$ ; the formula becomes $\int f'(x)[f(x)]^{-2}\,dx=-[f(x)]^{-1}+c$ , wherever $f(x)\ne0$ .

Watch out

$\int(x^2+1)^2\,dx$ does not have the same reverse-chain pattern: the factor proportional to $2x$ is missing. Dividing a guessed answer by $x$ is not a correction, because that quotient itself has a nonconstant derivative.

Change only the bracket power once more, from $3/2$ to $1/2$ .

Logarithmic integrals with absolute values

When the bracket power is $-1$ , the reverse-chain power rule gives way to a logarithm. Look for a numerator proportional to the derivative of the denominator.

For example, the derivative of $x^2-4$ is $2x$ . Consequently the ratio $2x/(x^2-4)$ is a logarithmic derivative. On a negative branch of the denominator, absolute-value signs keep the logarithm real.

Logarithmic recognition

On an interval where $f(x)\ne0$ ,

\int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+c.

If $f(x)>0$ throughout the stated domain, write $\ln f(x)$ . For $f(x)<0$ , the equivalent expression is $\ln(-f(x))$ .

Not in the formula booklet: learn it.

The absolute values do not bridge a zero of the denominator. Work on a connected interval where the original integrand is defined.

Worked example 3: a logarithmic derivative

Find $\int\frac{2x}{x^2-4}\,dx$ .

Plan. Match the numerator to the derivative of the denominator.

\frac{d}{dx}(x^2-4)=2x \quad\Longrightarrow\quad \int\frac{2x}{x^2-4}\,dx=\ln|x^2-4|+c.

Answer. $\ln|x^2-4|+c$ , on any interval avoiding $x=-2$ and $x=2$ .

Check. If $x^2-4>0$ , differentiation gives $2x/(x^2-4)$ . If $x^2-4<0$ , the answer is $\ln(4-x^2)$ , with derivative $-2x/(4-x^2)=2x/(x^2-4)$ .

Change only the numerator coefficient from $2$ to $5$ . A compensating constant is now required.

Worked example 4: numerator proportional to the derivative

Find $\int\frac{5x}{x^2-4}\,dx$ .

Plan. Write $5x=\frac52(2x)$ , keeping the denominator unchanged.

\int\frac{5x}{x^2-4}\,dx =\frac52\int\frac{2x}{x^2-4}\,dx =\frac52\ln|x^2-4|+c.

Answer. $\frac52\ln|x^2-4|+c$ .

Check. Differentiation gives $\frac52\frac{2x}{x^2-4}=5x/(x^2-4)$ .

There is no need to split this expression into fractions: the derivative is already visible in the numerator.

Watch out

$\int\frac{5x}{x^2-4}\,dx$ is not $5x\ln|x^2-4|+c$ . The numerator supplies a derivative factor; it is not a constant that can be left outside the integral.

Find the numerical ratio between the next numerator and the denominator's derivative.

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P3 5.2 - Recognising reverse chain rules

Recognising f-prime times a power of f

Logarithmic integrals with absolute values

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