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Anti-differentiate a power of a linear function

Start a set of 8
Step 3 of 9

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 6, 1 difficulty level, 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 inner expression is nonlinear, so the linear-inner rule does not apply. Here the exponent is two: expand the displayed polynomial and integrate each power

Difficulty levels

challengeDetermine a specific antiderivative of a linear-inner power using a function-value condition

Questions about this topic

Why is the inner coefficient in the denominator?+

Differentiating the linear bracket produces its inner coefficient. Dividing by that coefficient removes the extra 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 squared bracket in the displayed example. 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, Specialist Mathematics, chapter 6: Differential equations and rectilinear motion

In this chapter

6A Models, verification and direction fields: Form differential models, verify solutions and read direction fieldsExponential growth and decay rate modelsSeparate variables, impose initial values and retain solution domainsAnti-differentiate a power of a linear functionDifferentiate parametric curves and integrate vector motionSolve and interpret continuous logistic models6C Logistic growth and applications: Form differential models, verify solutions and read direction fieldsApproximate solutions with explicit Euler stepsBinary place value and algorithm tracesUse acceleration forms and differential equations in rectilinear motionAnalyse straight-line motion through displacement, velocity and acceleration