Integration using partial fractions and reverse chain ruleEdexcel A-Level Maths: Revision notes

Section 1

Integrals of the form f'(x)/f(x)

If the numerator is the derivative of the denominator, the integral is a logarithm: $\int\frac{f'(x)}{f(x)}\,\mathrm{d}x=\ln|f(x)|+c.$ This is the reverse of the chain rule applied to $\ln f(x)$ . Example: $\int\frac{6x^2}{2x^3+1}\,\mathrm{d}x=\ln|2x^3+1|+c$ because $\frac{\mathrm{d}}{\mathrm{d}x}(2x^3+1)=6x^2$ . If the numerator is only a constant multiple of $f'(x)$ , take the constant outside: $\int\frac{x}{x^2+5}\,\mathrm{d}x=\frac12\int\frac{2x}{x^2+5}\,\mathrm{d}x=\frac12\ln(x^2+5)+c$ .

Key termsreverse chain rulelogarithm

Common mistake

Integrating numerator and denominator separately. $\int\frac{x}{x^2+5}\,\mathrm{d}x$ is not $\frac{x^2/2}{x^3/3+5x}$ .

Exam tip

Differentiate the denominator first. If you get the numerator (or a constant multiple of it), the answer is a logarithm.

Section 2

Linear brackets: (ax + b)

For a linear expression $ax+b$ inside a power or a reciprocal, divide by $a$ : $\int\frac{1}{ax+b}\,\mathrm{d}x=\frac1a\ln|ax+b|+c,\qquad \int(ax+b)^n\,\mathrm{d}x=\frac{(ax+b)^{n+1}}{a(n+1)}+c\quad(n\ne-1).$ Examples: $\int\frac{2}{3x+5}\,\mathrm{d}x=\frac23\ln|3x+5|+c$ and $\int\frac{2}{(2x-1)^4}\,\mathrm{d}x=\int2(2x-1)^{-4}\,\mathrm{d}x=\frac{2(2x-1)^{-3}}{(-3)(2)}+c=-\frac{1}{3(2x-1)^3}+c$ . Check by differentiating: the chain rule brings back the factor $a$ , which is why you divide by it.

Key termslinear bracket

Common mistake

Using the power rule for $n=-1$ . $\int(3x+5)^{-1}\,\mathrm{d}x$ gives a logarithm, not $(3x+5)^0$ .

Section 3

Partial fractions

To integrate a proper fraction with a factorised denominator, split it into partial fractions first.

Distinct linear factors: $\frac{4x+11}{(x-1)(2x+3)}=\frac{A}{x-1}+\frac{B}{2x+3}$ .
A repeated factor: $\frac{px^2+qx+r}{(x+1)(2x-1)^2}=\frac{A}{x+1}+\frac{B}{2x-1}+\frac{C}{(2x-1)^2}$ . Multiply through by the denominator, then substitute convenient values of $x$ (the roots of each factor) or compare coefficients. Example: $4x+11=A(2x+3)+B(x-1)$ . Put $x=1$ : $A=3$ . Put $x=-\frac32$ : $B=-2$ .

Key termspartial fractionsrepeated factor

Common mistake

Leaving out the term $\frac{B}{2x-1}$ when a factor is repeated.

Exam tip

Check one value of $x$ (e.g. $x=0$ ) in the original and the split form.

Section 4

Integrating partial fractions

Integrate each partial fraction using the rules above: $\int\left(\frac{3}{x-1}-\frac{2}{2x+3}\right)\mathrm{d}x=3\ln|x-1|-\ln|2x+3|+c.$ A squared bracket in the denominator integrates by the reverse chain rule: $\int\frac{3}{(2x-1)^2}\,\mathrm{d}x=-\frac{3}{2(2x-1)}+c$ . Use the laws of logarithms to give a single logarithm when asked: $\int_2^5f(x)\,\mathrm{d}x=3\ln4-\ln13+\ln7=\ln\frac{448}{13}$ . Substitute the limits into the whole integrated expression, then subtract.

Key termsdefinite integral

Exam tip

Keep exact values ( $\ln2$ , $\ln5$ ) and combine with $a\ln b=\ln b^a$ and $\ln a-\ln b=\ln\frac ab$ .

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Integration using partial fractions and reverse chain rule

Two rational functions are defined for $x\ge0$ by $h(x)=\frac{2}{3x+5}$ and $k(x)=\frac{x}{x^2+5}$ .
Find the exact value of $\int_0^2 3k(x)\,\mathrm{d}x$ .2 marks
Let $f(x)=\frac{4x+11}{(x-1)(2x+3)}$ for $x>1$ .
Find the exact value of $\int_2^5 f(x)\,\mathrm{d}x$ , giving your answer as a single logarithm.2 marks
A curve $C$ has gradient $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2}{(2x-1)^4}$ for $x>\frac12$ , and passes through the point $(1,\,1)$ .
Find the equation of $C$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).