Find antiderivatives of rational and linear-composite powers
Reverse the power and chain rules, retain the constant and differentiate your result to check it.
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Section 1.8 includes rational and linear-composite powers except exponent −1, simple improper integrals, areas and volumes about either coordinate axis, including regions away from the rotation axis. Written practice is self-checked.
Before you begin
- Differentiate rational powers and linear-composite powers.
- Rewrite roots and reciprocals using indices.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the derivative of x³ + 7?
What you will learn
- Integrate constant multiples, sums and differences of rational powers with exponent other than −1.
- Integrate (ax + b)ⁿ with the correct inner scale and check the real domain.
Reverse a derivative and keep the family of answers
An antiderivative F of f satisfies F′(x) = f(x). Since differentiating a constant gives zero, ∫f(x) dx means F(x) + C, a family with an arbitrary constant. For rational n ≠ −1, ∫xⁿ dx = x^(n + 1)/(n + 1) + C on intervals where the real expressions are valid. Increase the exponent first, then divide by the new exponent; integrate sums and differences term by term.
For x > 0, 3x² − 4√x + 6/x² is 3x² − 4x^(1/2) + 6x^(−2). Its antiderivative is x³ − (8/3)x^(3/2) − 6/x + C. Differentiating gives the original expression, including the positive reciprocal-square term. Exponent −1 is excluded from this power formula because n + 1 would be zero; its logarithmic antiderivative belongs to later content.
Undo the inner scale for a linear composite
For a ≠ 0 and n ≠ −1, ∫(ax + b)ⁿ dx = (ax + b)^(n + 1)/[a(n + 1)] + C. Differentiation would multiply by a, so integration divides by it. For example, ∫(3x + 2)⁴ dx = (3x + 2)⁵/15 + C. This shortcut applies to a linear inner expression; using it unchanged for an arbitrary quadratic inner expression omits a changing factor.
State a valid real interval when roots or reciprocals occur. For instance ∫(2x + 1)^(−2) dx = −1/[2(2x + 1)] + C on either side of x = −1/2, with potentially different constants on the separate intervals. Do not integrate across an undefined input as though the function were continuous there. If a = 0, the integrand is a constant wherever defined; integrate that constant directly rather than divide by zero.
Pause and explain
Which is an antiderivative of (3x + 2)⁴?
Worked example
For x > 0, integrate 3x² − 4√x + 6/x² and check your result.
Show the worked solution
- Rewrite the integrand as 3x² − 4x^(1/2) + 6x^(−2).
- Increase each power by 1 and divide by its new exponent to obtain x³ − (8/3)x^(3/2) − 6/x + C.
- Differentiate: 3x² − 4x^(1/2) + 6x^(−2), recovering every original term.
Answer x³ − (8/3)x^(3/2) − 6/x + C, for x > 0.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Integrate 6x² − 4x + 3 and verify by differentiation.
Give me a hint
Treat each term separately and add one arbitrary constant at the end.
Compare my reasoning
- ∫6x² dx = 2x³ and ∫−4x dx = −2x².
- The constant 3 integrates to 3x, giving 2x³ − 2x² + 3x + C.
- Differentiating term by term gives 6x² − 4x + 3.
2x³ − 2x² + 3x + C.
Look for these in your work
- Divide each coefficient by the increased exponent.
- Recover the original integrand in your check.
For x > 0, integrate 2√x − 5/x² and check both signs.
Give me a hint
Write powers 1/2 and −2 before applying the rule.
Compare my reasoning
- Rewrite as 2x^(1/2) − 5x^(−2).
- The antiderivative is (4/3)x^(3/2) + 5x^(−1) + C.
- Its derivative is 2√x − 5/x², so the positive reciprocal term in the answer is correct.
(4/3)x^(3/2) + 5/x + C, for x > 0.
Look for these in your work
- Handle a negative new exponent with its sign.
- State the real interval and retain C.
A supplied flow model is q(t) = 3(2t + 1)² litres per minute for 0 ≤ t ≤ 2. Find an antiderivative F(t), then use F(2) − F(0) to calculate the added volume.
Give me a hint
Reverse the inner factor 2, then use the change in accumulated volume so C cancels.
Compare my reasoning
- An antiderivative is F(t) = (2t + 1)³/2 + C, whose derivative is 3(2t + 1)².
- F(2) − F(0) = (5³ − 1³)/2 = 62.
- The added volume is 62 litres; the same constant cancels between the two endpoint values.
F(t) = (2t + 1)³/2 + C; added volume 62 litres.
Look for these in your work
- Reverse both the exponent and inner gradient.
- Give accumulated volume units and explain cancellation of C.
Common mistakes
- Divide by the new exponent, not the old one.
- A linear inner expression introduces an additional divisor a.
- Keep C for an indefinite integral and exclude exponent −1 from this power formula.
What can you explain now?
Integrate (3x + 2)⁴ with respect to x.
Compare with the explanation
(3x + 2)⁵/15 + C.
Differentiation multiplies by 5 and by the inner gradient 3; dividing by 15 reverses both.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Explain how differentiating an antiderivative checks its coefficient, inner scale and signs. State why exponent −1 needs a different rule.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.