Mc27, Mc28 - Integration by substitution | OCR A-Level Mathematics B (MEI)

Recognise reverse-chain integrals and choose one useful change of variable in less obvious cases. Rewrite every factor, preserve domains and convert definite bounds consistently.

Recognise a function and its derivative

How can an exponential containing a square be easier to integrate than it looks? In $\int x\mathrm e^{x^2}\,\mathrm dx$ , the outside $x$ almost matches the inner derivative $2x$ . You need a coefficient adjustment, not an expansion.

You already know

The chain rule differentiates an outer function and multiplies by the inner derivative. Reverse it when that derivative, or a constant multiple, appears in the integrand.

With $u=g(x)$ , the equation $\mathrm du=g'(x)\,\mathrm dx$ describes the differential conversion. It does not mean that every $\mathrm dx$ can simply become $\mathrm du$ .

Reverse-chain power integral

For $n\ne-1$ , on an interval where all expressions are real and defined,

\int g'(x)\bigl(g(x)\bigr)^n\,\mathrm dx =\frac{\bigl(g(x)\bigr)^{n+1}}{n+1}+C.

In the formula sheet.

A recognised Mc27 integral can be written by inspection; the substitution below explains the coefficient. Show it when the question requests it.

Worked example 1: an exponential inner function

Question. Find $\int x\mathrm e^{x^2}\,\mathrm dx$ .

Plan. Put $u=x^2$ , so $x\,\mathrm dx=\mathrm du/2$ .

\int x\mathrm e^{x^2}\,\mathrm dx =\frac12\int\mathrm e^u\,\mathrm du =\frac12\mathrm e^{x^2}+C.

Why the half? Differentiating $x^2$ produces twice the outside factor.

Answer. $\mathrm e^{x^2}/2+C$ .

Check. The derivative is $\frac12\mathrm e^{x^2}(2x)$ , exactly the integrand.

Change only the exponent from $x^2$ to $3x^2$ . The differential multiplier changes with it.

Worked example 2: adjusting an inner coefficient

Question. Find $\int x\mathrm e^{3x^2}\,\mathrm dx$ .

Plan. Use $u=3x^2$ , hence $x\,\mathrm dx=\mathrm du/6$ .

\int x\mathrm e^{3x^2}\,\mathrm dx =\frac16\int\mathrm e^u\,\mathrm du =\frac16\mathrm e^{3x^2}+C.

Why not keep the old half? The inner derivative is now $6x$ .

Answer. $\mathrm e^{3x^2}/6+C$ .

Check. Differentiation cancels the denominator $6$ .

The same structure yields $\int x(x^2+4)^3\,\mathrm dx=(x^2+4)^4/8+C$ and $\int(2x+1)^5\,\mathrm dx=(2x+1)^6/12+C$ . An expansion can turn a one-line calculation into several unnecessary ones: recognition is a real mathematical shortcut.

Watch out

Wrong: raise the bracket's power without checking its derivative. Right: differentiate your proposed primitive once to check the multiplier.

Change the coefficient inside the exponential again.

A derivative over its function gives a logarithm

A numerator matching the derivative of a denominator signals a logarithm, not the ordinary power rule. The power $-1$ is the dividing line.

Logarithmic derivative

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

\int\frac{k g'(x)}{g(x)}\,\mathrm dx=k\ln|g(x)|+C.

In the formula sheet.

Modulus permits negative values of the denominator, but never its zeros. Constants may be chosen separately on disconnected intervals.

Worked example 3: a cubic denominator

Question. Find $\int(x^2+2x)/(x^3+3x^2-5)\,\mathrm dx$ .

Plan. Compare the numerator with the derivative of $g(x)=x^3+3x^2-5$ .

\begin{aligned} g'(x)&=3x^2+6x=3(x^2+2x) &&\text{identify the ratio}\\ I&=\frac13\int\frac{g'(x)}{g(x)}\,\mathrm dx =\frac13\ln|x^3+3x^2-5|+C. \end{aligned}

Why one third? The denominator's derivative is three times the numerator.

Answer. $\frac13\ln|x^3+3x^2-5|+C$ , on intervals excluding its zeros.

Check. Differentiate the logarithm and cancel the factor $3$ .

Change only the numerator's multiplier from $1$ to $2$ ; the domain is unchanged.

Worked example 4: the numerator multiplier

Question. Find $\int(2x^2+4x)/(x^3+3x^2-5)\,\mathrm dx$ .

Plan. Rewrite the numerator as $2g'(x)/3$ .

I=\frac23\int\frac{g'(x)}{g(x)}\,\mathrm dx =\frac23\ln|x^3+3x^2-5|+C.

Why keep the same denominator? Only its derivative ratio changed; the same logarithm still fits.

Answer. $\frac23\ln|x^3+3x^2-5|+C$ .

Check. Differentiation gives $2(x^2+2x)/(x^3+3x^2-5)$ .

A quadratic example is $\int(2x+1)/(x^2+x-3)\,\mathrm dx=\ln|x^2+x-3|+C$ . Trigonometry can hide the pattern: $\tan x=\sin x/\cos x$ gives $-\ln|\cos x|+C$ , in radians and away from zeros of cosine. The last result follows from the supplied general formula; it is not separately printed.

Watch out

Wrong: a logarithm automatically has every real input. Right: exclude zeros of its original denominator; omit modulus only when positivity has been established.

Choose a quadratic whose positivity can be proved exactly.

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Mc27, Mc28 - Integration by substitution

Recognise a function and its derivative

A derivative over its function gives a logarithm

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