Indefinite integrationEdexcel International A Level Maths: Revision notes

Section 1

Integration as the reverse of differentiation

Integration reverses differentiation. If $\frac{\mathrm{d}y}{\mathrm{d}x}=f(x)$ then $y=\int f(x)\,\mathrm{d}x$ , which is the indefinite integral of $f(x)$ with respect to $x$ . Because a constant differentiates to zero, functions such as $x^2$ , $x^2+5$ and $x^2-7$ all have the same derivative $2x$ . So every indefinite integral must include an arbitrary constant of integration $c$ : $\int2x\,\mathrm{d}x=x^2+c.$ You can always check an integral by differentiating it.

Key termsindefinite integralconstant of integration

Common mistake

Leaving out $+c$ . The answer is a family of functions, not a single one.

Exam tip

Differentiate your answer to check that it gives back the integrand.

Section 2

Integrating powers of x

For any rational $n\neq-1$ , $\int x^n\,\mathrm{d}x=\frac{x^{n+1}}{n+1}+c.$ Add one to the power, then divide by the new power. A constant multiple stays in place, and sums and differences are integrated term by term: $\int ax^n\,\mathrm{d}x=\frac{ax^{n+1}}{n+1}+c.$ A constant integrates as $\int k\,\mathrm{d}x=kx+c$ . Example: $\int(6x^2-4x+5)\,\mathrm{d}x=2x^3-2x^2+5x+c$ . The case $n=-1$ ( $\int\frac1x\,\mathrm{d}x$ ) is excluded here, because the rule would need division by zero.

Key termsintegrand

Common mistake

Not dividing by the new power: $\int6x^2\,\mathrm{d}x\neq6x^3$ .

Common mistake

Integrating a constant $5$ as $5$ instead of $5x$ .

Section 3

Fractional and negative powers

Rewrite roots and reciprocals as powers before integrating, then use the same rule. $\int x^{-3}\,\mathrm{d}x=\frac{x^{-2}}{-2}+c=-\frac12x^{-2}+c,\qquad\int x^{\frac12}\,\mathrm{d}x=\frac{x^{\frac32}}{\frac32}+c=\frac23x^{\frac32}+c.$ Dividing by a fraction means multiplying by its reciprocal. Example: $\int\left(4\sqrt{x}-\frac{6}{x^3}\right)\mathrm{d}x=\int\left(4x^{\frac12}-6x^{-3}\right)\mathrm{d}x=\frac83x^{\frac32}+3x^{-2}+c$ .

Common mistake

Subtracting one from the power when the power is negative. For $x^{-3}$ you add one to get $x^{-2}$ , then divide by $-2$ .

Section 4

Simplifying before integrating

Integration has no product or quotient rule at this level. If the integrand is a product of brackets or a fraction with a single term on the bottom, expand and divide each term first. Example: $\frac{(2x+1)^2}{\sqrt{x}}=\frac{4x^2+4x+1}{x^{\frac12}}=4x^{\frac32}+4x^{\frac12}+x^{-\frac12}$ , so $\int\frac{(2x+1)^2}{\sqrt{x}}\,\mathrm{d}x=\frac85x^{\frac52}+\frac83x^{\frac32}+2x^{\frac12}+c.$ In the same way $\frac{x^3-2x}{\sqrt x}=x^{\frac52}-2x^{\frac12}$ . An expression such as $\frac{(x+2)^2}{x}$ expands to $x+4+4x^{-1}$ . Its $x^{-1}$ term needs the excluded case $n=-1$ , so choose forms like $\frac{(x+2)^2}{\sqrt x}$ for practice.

Key termsexpand

Common mistake

Integrating a product factor by factor. $\int x(x+3)\,\mathrm{d}x$ is not $\frac{x^2}{2}\left(\frac{x^2}{2}+3x\right)$ ; expand to $x^2+3x$ first.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Indefinite integration

Let $g(x)=6x^2-4x+5$ .
Find $\int\left(g(x)+\frac{4}{x^3}\right)\mathrm{d}x$ .2 marks
A function is defined by $f(x)=4\sqrt{x}-\dfrac{6}{x^3}$ for $x>0$ .
Find $\int x^3f(x)\,\mathrm{d}x$ .2 marks
The function $f$ is defined by $f(x)=\dfrac{(2x+1)^2}{\sqrt{x}}$ for $x>0$ .
Show that $f(x)=4x^{\frac32}+4x^{\frac12}+x^{-\frac12}$ .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).