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Anti-differentiate a power of a linear function

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The linear-inner power rule

For a power of a linear expression, increase the exponent by one and divide the outer coefficient by the inner coefficient times the new exponent. For a polynomial bracket raised to a nonnegative integer power, expansion gives a finite polynomial that can be integrated term by term. This is not a general method for nonlinear reciprocals or roots

\[\int c\left(ax+b\right)^n\,dx=\frac{c}{a(n+1)}\left(ax+b\right)^{n+1}+C,\quad a\ne0,\;n\ne-1\]
  1. 1Check the inner expression Use this rule only when the bracket has the form ax+b with a non-zero constant a. \(\displaystyle \left(3x-2\right)^4\text{ has a linear inner expression}\)
  2. 2Increase the exponent Replace n with n+1 in the primitive \(\displaystyle \int\left(2x+1\right)^3\,dx=\frac{\left(2x+1\right)^4}{8}+C\)
  3. 3Divide by the full factor Divide the outer coefficient by a(n+1), not by n or n+1 alone \(\displaystyle \int6\left(2x+1\right)^3\,dx=\frac{6}{2(4)}\left(2x+1\right)^4+C\)
  4. 4Expand a polynomial power When a polynomial bracket has a nonnegative integer exponent, expand it before applying the ordinary power rule term by term \(\displaystyle \int\left(x^2+1\right)^2\,dx=\frac{x^5}{5}+\frac{2x^3}{3}+x+C\)
  5. 5Use a valid interval Work on a connected interval where the real integrand and primitive are defined. For a negative integer power, exclude ax+b=0; the constants on the two sides can differ. The exponent −1 needs the logarithmic rule, not division by n+1. \(\displaystyle \int3(2x+1)^{-2}\,dx=-\frac{3}{2(2x+1)}+C,\quad x\ne-\frac{1}{2}\)
  6. 6Interpret C in the displayed form Setting C=0 removes the additive constant outside the displayed bracket. It does not usually mean F(0)=0 or that the expanded polynomial has no constant term \(\displaystyle F(x)=\frac{3}{4}(2x-1)^4+C,\quad C=0\implies F(0)=\frac{3}{4}\)
Year 12, VCE Units 3 & 4, Chapter 7, 4 difficulty levels, No calculatorBuild a worksheet

The linear-inner power rule

For a power of a linear expression, increase the exponent by one and divide the outer coefficient by the inner coefficient times the new exponent. For a polynomial bracket raised to a nonnegative integer power, expansion gives a finite polynomial that can be integrated term by term. This is not a general method for nonlinear reciprocals or roots

\[\int c\left(ax+b\right)^n\,dx=\frac{c}{a(n+1)}\left(ax+b\right)^{n+1}+C,\quad a\ne0,\;n\ne-1\]
  1. Check the inner expression Use this rule only when the bracket has the form ax+b with a non-zero constant a. \(\displaystyle \left(3x-2\right)^4\text{ has a linear inner expression}\)
  2. Increase the exponent Replace n with n+1 in the primitive \(\displaystyle \int\left(2x+1\right)^3\,dx=\frac{\left(2x+1\right)^4}{8}+C\)
  3. Divide by the full factor Divide the outer coefficient by a(n+1), not by n or n+1 alone \(\displaystyle \int6\left(2x+1\right)^3\,dx=\frac{6}{2(4)}\left(2x+1\right)^4+C\)
  4. Expand a polynomial power When a polynomial bracket has a nonnegative integer exponent, expand it before applying the ordinary power rule term by term \(\displaystyle \int\left(x^2+1\right)^2\,dx=\frac{x^5}{5}+\frac{2x^3}{3}+x+C\)
  5. Use a valid interval Work on a connected interval where the real integrand and primitive are defined. For a negative integer power, exclude ax+b=0; the constants on the two sides can differ. The exponent −1 needs the logarithmic rule, not division by n+1. \(\displaystyle \int3(2x+1)^{-2}\,dx=-\frac{3}{2(2x+1)}+C,\quad x\ne-\frac{1}{2}\)
  6. Interpret C in the displayed form Setting C=0 removes the additive constant outside the displayed bracket. It does not usually mean F(0)=0 or that the expanded polynomial has no constant term \(\displaystyle F(x)=\frac{3}{4}(2x-1)^4+C,\quad C=0\implies F(0)=\frac{3}{4}\)

Worked example

Find the complete family of anti-derivatives of the function below

\[\int4\left(3x+2\right)^2\,dx\]
  1. \(\displaystyle a=3,\quad b=2,\quad c=4,\quad n=2\)The inner expression is linear, with inner coefficient 3 and outer coefficient 4.
  2. \(\displaystyle n+1=2+1=3\)Increase the power by one
  3. \(\displaystyle a(n+1)=3(3)=9,\quad\frac{c}{a(n+1)}=\frac{4}{9}\)Divide the outer coefficient by the inner coefficient times the new exponent
  4. \(\displaystyle \int4\left(3x+2\right)^2\,dx=\frac{4}{9}\left(3x+2\right)^3+C\)Substitute the new exponent and coefficient into the rule and include one arbitrary constant

Common mistakes

Omitting the inner coefficient
✕\(\displaystyle \int6\left(2x-1\right)^3\,dx=\frac{3}{2}\left(2x-1\right)^4+C\)✓\(\displaystyle \int6\left(2x-1\right)^3\,dx=\frac{3}{4}\left(2x-1\right)^4+C\)
The coefficient must be divided by both the inner coefficient 2 and the new exponent 4.
Leaving the exponent unchanged
✕\(\displaystyle \int6\left(2x-1\right)^3\,dx=\frac{3}{4}\left(2x-1\right)^3+C\)✓\(\displaystyle \int6\left(2x-1\right)^3\,dx=\frac{3}{4}\left(2x-1\right)^4+C\)
The power rule increases the exponent from 3 to 4.
Using the rule on a non-linear inner expression
✕\(\displaystyle \int\left(x^2+1\right)^2\,dx=\frac{1}{3}\left(x^2+1\right)^3+C\)✓\(\displaystyle \int\left(x^2+1\right)^2\,dx=\frac{x^5}{5}+\frac{2x^3}{3}+x+C\)
The displayed inner expression is nonlinear, so the linear-inner rule does not apply. Here the exponent is 2: expand this finite polynomial and integrate each power

Difficulty levels

introAnti-differentiate one power of a linear expression with whole coefficients
standardHandle a signed inner constant, a constant multiple or a value with the integration constant set to zero in the displayed primitive
challengeExpand a polynomial bracket raised to a positive integer power, or determine a specific member using a condition
examChoose the correct primitive from distractors involving the coefficient, exponent and chain factor

Questions about this topic

Why is the inner coefficient in the denominator?+

Differentiating the linear expression produces its inner coefficient as an extra factor. Dividing by that coefficient removes the factor from the derivative of the primitive

Why must the exponent increase by one?+

Differentiation lowers a power by one, so anti-differentiation reverses that change by increasing the exponent by one

When should the bracket be expanded first?+

Expand a polynomial bracket raised to a nonnegative integer power, such as the polynomial power displayed above. A nonlinear reciprocal or root does not generally expand to a finite polynomial; the linear-inner rule cannot be used by treating a variable inner derivative as a constant

Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 7: Antiderivatives

In this chapter

7A Antiderivatives and arbitrary constants: Find complete families of polynomial anti-derivatives7B Power rule and integrand preparation: Find complete families of polynomial anti-derivatives7B Power rule and integrand preparation: Recognise equivalent polynomial formsSimplify and combine algebraic fractionsAnti-differentiate a power of a linear function7C Integrating powers of linear functions: Recognise equivalent polynomial forms7C Integrating powers of linear functions: Solve a linear equationAnti-differentiate exponential and circular functionsAnti-differentiate by recognition from a supplied derivativeDetermine a polynomial from its derivative and a boundary conditionInfer primitive curves from a gradient functionReconstruct quantities from their ratesEvaluate and determine function rules7E Antiderivative families and applications: Solve a linear equationRound to decimal places and significant figuresScale a proportional quantity with the unitary method