Pure 8.2 - Integrating standard functions | Edexcel A-Level Mathematics

How can you recover a curve when its derivative contains powers, fractions and trigonometric functions together? You will build a small set of integration rules, rewrite unfamiliar-looking expressions into those forms, and use a point to identify the required curve. Choosing the form before calculating keeps the work manageable.

Integration as a reverse process

A curve has gradient $6x^2+4\sec^2 2x$ and passes through $(0,-1)$ . What is its equation? Start with the easier term: $\frac{\mathrm{d}}{\mathrm{d}x}(2x^3)=6x^2$ . Recovering the other term needs the same backward thinking; you will put the pieces together in the last part.

You already know

An indefinite integral describes antiderivatives and needs $+C$ . Differentiate your proposed answer to check it; a point on the original curve determines $C$ .

For $x^n$ , try a multiple of $x^{n+1}$ . Differentiating multiplies its coefficient by $n+1$ , so divide by that number first.

Power rule for integration

For rational $n\ne-1$ , on an interval where the powers are real and the integrand is continuous,

\int x^n\,\mathrm{d}x=\frac{x^{n+1}}{n+1}+C

Integrate sums and differences term by term and keep constant multipliers.

Not in the formula booklet: learn it.

The restriction matters: $n=-1$ would require division by zero. That case needs a logarithm. For fractional powers, check the real domain; for negative powers, do not cross a singularity.

Worked example 1: a sum of powers

Question. Find $\int(6x^2-4x^{1/3}+5)\,\mathrm{d}x$ .

Plan. Increase each power by one, divide by the new power, and combine the antiderivatives.

\begin{aligned} \int(6x^2-4x^{1/3}+5)\,\mathrm{d}x &=\frac{6x^3}{3}-\frac{4x^{4/3}}{4/3}+5x+C && \text{apply the power rule}\\ &=2x^3-3x^{4/3}+5x+C && \text{simplify coefficients} \end{aligned}

Why this step? Dividing by $4/3$ means multiplying by $3/4$ , so the coefficient of $x^{4/3}$ is $-3$ .

Answer. $2x^3-3x^{4/3}+5x+C$ .

Check. Its derivative is $6x^2-3(4/3)x^{1/3}+5=6x^2-4x^{1/3}+5$ , including at $x=0$ by the derivative limit.

Change only the fractional exponent from $1/3$ to $-1/2$ . The coefficient and the other terms stay fixed, but the real domain is now $x>0$ .

Worked example 2: a negative fractional power

Question. For $x>0$ , find $\int(6x^2-4x^{-1/2}+5)\,\mathrm{d}x$ .

Plan. Apply the same power rule with the new exponent.

\begin{aligned} \int(6x^2-4x^{-1/2}+5)\,\mathrm{d}x &=2x^3-\frac{4x^{1/2}}{1/2}+5x+C && \text{new power }1/2\\ &=2x^3-8\sqrt{x}+5x+C && \text{divide by }1/2 \end{aligned}

Why this step? A negative exponent is allowed; only $-1$ is excluded from the power rule.

Answer. $2x^3-8\sqrt{x}+5x+C$ , for $x>0$ .

Check. The derivative of $-8x^{1/2}$ is $-4x^{-1/2}$ , so all three terms return correctly.

A useful surprise is that the exceptional power $x^{-1}$ leads to a logarithm, while the nearby power $x^{-2}$ leads to another power. One exponent changes the family of function you need.

Watch out

Wrong: $\int x^{-1/2}\,\mathrm{d}x=-2x^{-3/2}$ . That differentiates the power. Right: increase the power first, giving $2x^{1/2}+C$ .

Try the same method with another negative fractional exponent coefficient.

Rewrite rational expressions first

A fraction may conceal several standard powers. For example, dividing each numerator term by $x^2$ gives $\frac{x+2}{x^2}=x^{-1}+2x^{-2}$ , for $x\ne0$ .

Rational function

A rational function is a quotient of polynomials, on a domain excluding zeros of its denominator. For example, $\frac{x+2}{x^2}$ is rational; $\sqrt{x}$ is not a quotient of polynomials.

The $x^{-1}$ term has a different reverse derivative. On either interval $x>0$ or $x<0$ , differentiating $\ln|x|$ gives $1/x$ .

Reciprocal integral

On an interval not containing zero,

\int\frac1x\,\mathrm{d}x=\ln|x|+C

When $x>0$ , this is $\ln x+C$ .

Not in the formula booklet: learn it.

A connected interval matters: the constants on the positive and negative sides need not be equal. Do not use this formula to evaluate an ordinary definite integral across zero.

Worked example 3: separating the fraction

Question. Find $\int\frac{x+2}{x^2}\,\mathrm{d}x$ , for $x\ne0$ .

Plan. Split the fraction into a reciprocal term and a power term.

\begin{aligned} \frac{x+2}{x^2}&=x^{-1}+2x^{-2} && \text{divide each term}\\ \int\frac{x+2}{x^2}\,\mathrm{d}x&=\ln|x|-2x^{-1}+C && \text{use the two rules} \end{aligned}

Why this step? Only the $x^{-1}$ term becomes a logarithm; $x^{-2}$ still uses the power rule.

Answer. $\ln|x|-\frac2x+C$ , on either side of zero separately.

Check. Differentiation gives $1/x+2/x^2=(x+2)/x^2$ .

Change only the numerator from $x+2$ to $(x+2)^2$ . Expanding before dividing reveals an additional constant term.

Worked example 4: a squared numerator

Question. Find $\int\frac{(x+2)^2}{x^2}\,\mathrm{d}x$ , for $x\ne0$ .

Plan. Expand the numerator, divide term by term, then integrate.

\begin{aligned} \frac{(x+2)^2}{x^2}&=\frac{x^2+4x+4}{x^2} && \text{expand}\\ &=1+\frac4x+4x^{-2} && \text{divide}\\ \int\frac{(x+2)^2}{x^2}\,\mathrm{d}x&=x+4\ln|x|-\frac4x+C && \text{integrate each term} \end{aligned}

Why this step? Expanding makes every term match a known rule; no special rule for integrating a quotient is needed.

Answer. $x+4\ln|x|-4/x+C$ .

Check. Its derivative is $1+4/x+4/x^2=(x+2)^2/x^2$ .

Other fractional denominators work the same way. For $x>0$ , $\frac{(x+2)^2}{x^{1/2}}=x^{3/2}+4x^{1/2}+4x^{-1/2}$ , whose antiderivative is $\frac25x^{5/2}+\frac83x^{3/2}+8x^{1/2}+C$ . Differentiating checks all three coefficients.

Watch out

Wrong: $\int\frac{f(x)}{g(x)}\,\mathrm{d}x=\frac{\int f(x)\,\mathrm{d}x}{\int g(x)\,\mathrm{d}x}$ . Integration has no such quotient rule. Right: rewrite into recognised forms, as above.

Change the constant inside the squared numerator and repeat the same three steps.

Standard functions with kx

Differentiating $\sin5x$ gives $5\cos5x$ . To reverse that multiplication by $5$ , integrate $\cos5x$ as $\frac15\sin5x$ . The same compensation works whenever the inner function is $kx$ .

Basic integrals with a linear input

For constant $k\ne0$ , with angles in radians,

\begin{aligned} \int e^{kx}\,\mathrm{d}x&=\frac1k e^{kx}+C\\ \int\sin kx\,\mathrm{d}x&=-\frac1k\cos kx+C\\ \int\cos kx\,\mathrm{d}x&=\frac1k\sin kx+C \end{aligned}

Not in the formula booklet: learn it.

For $k=0$ , the integrands become constants and must be integrated as constants; the displayed divisions by $k$ do not apply. Reciprocal trigonometric functions also require intervals avoiding their singularities.

Supplied trigonometric integrals

For $k\ne0$ , in radians and on an interval where the integrand is continuous,

\begin{aligned} \int\sec^2 kx\,\mathrm{d}x&=\frac1k\tan kx+C\\ \int\tan kx\,\mathrm{d}x&=\frac1k\ln|\sec kx|+C \end{aligned}

The second answer is equivalently $-\frac1k\ln|\cos kx|+C$ .

In the formula booklet.

For example, $\int\sin5x\,\mathrm{d}x=-\frac15\cos5x+C$ , because differentiating the answer gives $-\frac15(-5\sin5x)=\sin5x$ . Also $\int e^{x/5}\,\mathrm{d}x=5e^{x/5}+C$ : here $k=1/5$ , so its reciprocal is $5$ .

Worked example 5: a mixed integral

Question. Find $\int\left(4e^{2x}-3\sin5x+6\sec^2 3x+\frac1{2x}\right)\,\mathrm{d}x$ .

Plan. Apply the appropriate standard form separately to each term.

\begin{aligned} \int4e^{2x}\,\mathrm{d}x&=2e^{2x} && \text{divide by }2\\ \int-3\sin5x\,\mathrm{d}x&=\frac35\cos5x && \text{two negative factors}\\ \int6\sec^2 3x\,\mathrm{d}x&=2\tan3x && \text{divide by }3\\ \int\frac1{2x}\,\mathrm{d}x&=\frac12\ln|x| && \text{constant multiple of }1/x \end{aligned}

Why this step? $1/(2x)=\frac12(1/x)$ ; its coefficient belongs outside the reciprocal integral.

Answer. $2e^{2x}+\frac35\cos5x+2\tan3x+\frac12\ln|x|+C$ , on an interval avoiding $x=0$ and the poles of $\sec3x$ .

Check. Differentiating the four terms gives multipliers $2(2)$ , $(3/5)(-5)$ , $2(3)$ and $1/2$ , recovering the integrand.

Change only the exponential's inner coefficient from $2$ to $-2$ . Its antiderivative changes sign; the other terms do not change.

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Last Accessed

Pure Mathematics (Papers 1 and 2)

Statistics (Paper 3, Section A)

Mechanics (Paper 3, Section B)

Pure 8.2 - Integrating standard functions

Integration as a reverse process

Rewrite rational expressions first

Standard functions with kx

Continue this lesson.

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