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Calculus

Calculus

Integration: powers, 1/x and (ax + b)ⁿ

Integration and definite integrals (14.10-14.12) Paper 1 + 2 7 min read

On this page
  1. In this note
  2. Formula bank
  3. Integration undoes differentiation
  4. Powers of x
  5. A bracket to a power: (ax + b)ⁿ
  6. Ways they test it
  7. From the past papers
  8. Mistakes to avoid
  9. Summary
  10. Practice
  11. Can I do this?
  12. Quiz

In this note

Syllabus 14.10-14.12: the basic integrals
  • 14.10Understand integration as the reverse of differentiation, and why every answer needs + c.
  • 14.11Integrate sums of powers of x, including \(\dfrac1x\).
  • 14.12Integrate \((ax+b)^{n}\) for any rational n, including \(\dfrac{1}{ax+b}\).

Before you start: index laws (\(\sqrt x=x^{1/2}\), \(\tfrac1{x^3}=x^{-3}\)) and differentiating powers and the chain rule. Next: Integrating e and trig.

Formula bank

PowersLearn it\(\displaystyle\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+c\)add 1 to the power, divide by the NEW power. Any n except −1.Use it when every power of x, roots, 1/x²…
A constantLearn it\(\displaystyle\int k\,dx=kx+c\)Use it when a number on its own.
One over xLearn it\(\displaystyle\int \frac1x\,dx=\ln x+c\)the n = −1 case.Use it when \(\tfrac1x\), \(\tfrac5x\).
Bracket to a powerLearn it\(\displaystyle\int (ax+b)^{n}\,dx=\frac{(ax+b)^{n+1}}{a(n+1)}+c\)raise the power, divide by the new power AND by a.Use it when \((3x-1)^5\), \(\sqrt{4x+1}\).
One over a bracketLearn it\(\displaystyle\int \frac{1}{ax+b}\,dx=\frac1a\ln(ax+b)+c\)Use it when \(\tfrac{4}{2x+5}\).

No calculus formulas are printed in the exam: learn all of these.

Integration undoes differentiation

Differentiating \(x^{3}\) gives \(3x^{2}\). So integrating \(3x^{2}\) takes you back: \(\int 3x^{2}\,dx=x^{3}+c\). Why the \(+c\)? \(x^{3}+5\) and \(x^{3}-2\) also differentiate to \(3x^{2}\): the constant disappears when you differentiate, so when you integrate you don't know it. Write \(+c\) on EVERY indefinite integral (one with no limits).

General methodHow to check any integral

  1. Differentiate your answer.
  2. You should get back exactly what you started with. If not, fix the number in front.

Powers of x

General methodHow to integrate any sum of powers

  1. Rewrite every term as a number × a power of x: \(\sqrt x=x^{1/2}\), \(\tfrac{4}{x^{3}}=4x^{-3}\). Multiply out brackets; split a fraction over a single term into separate terms.
  2. Integrate term by term: add 1 to the power, divide by the new power.
  3. A term in \(x^{-1}\) (that is \(\tfrac{k}{x}\)) becomes \(k\ln x\).
  4. Add + c. Tidy up (negative powers can stay as they are).

Example Find \(\displaystyle\int\left(6x^{2}-\frac{4}{x^{3}}+3\sqrt x\right)dx\).

Rewrite, then raise each power by one.

  1. \(=\int(6x^{2}-4x^{-3}+3x^{1/2})\,dx\)
  2. \(=\dfrac{6x^{3}}{3}-\dfrac{4x^{-2}}{-2}+\dfrac{3x^{3/2}}{3/2}+c\)

Answer \(2x^{3}+2x^{-2}+2x^{3/2}+c\)

Example Find \(\displaystyle\int\frac{(2x+1)^{2}}{x}\,dx\).

Expand, then split over x.

  1. \(\dfrac{4x^{2}+4x+1}{x}=4x+4+\dfrac1x\)

Answer \(2x^{2}+4x+\ln x+c\)

Try it yourself

1
Find \(\int(4x^{3}-6x+5)\,dx\). (Type x^4 for \(x^4\); end with + c.)
Show answer

Answer \(x^{4}-3x^{2}+5x+c\)

2
Find \(\displaystyle\int\frac{x^{2}+3}{\sqrt x}\,dx\).
Hint
Split: x^(3/2) + 3x^(−1/2).
Show answer

Answer \(\tfrac25x^{5/2}+6x^{1/2}+c\)

A bracket to a power: (ax + b)ⁿ

Don't expand \((3x-1)^{5}\). Think of the chain rule backwards: differentiating \((3x-1)^{6}\) gives \(6\times3\times(3x-1)^{5}\). So to undo it, raise the power by one and divide by the new power AND by the 3 inside.

General methodHow to integrate (ax + b)ⁿ

  1. Keep the bracket. Add 1 to the power.
  2. Divide by the new power and by a (the number in front of x).
  3. Power −1, i.e. \(\dfrac{k}{ax+b}\)? Then it is \(\dfrac{k}{a}\ln(ax+b)\).
  4. Add + c; check by differentiating (chain rule).

Example Find \(\int(3x-1)^{5}\,dx\) and \(\displaystyle\int\frac{1}{\sqrt{4x+1}}\,dx\).

Raise the power, divide by the new power and by a.

  1. \(\dfrac{(3x-1)^{6}}{6\times3}=\dfrac{(3x-1)^{6}}{18}\)
  2. \((4x+1)^{-1/2}\to\dfrac{(4x+1)^{1/2}}{\tfrac12\times4}=\tfrac12(4x+1)^{1/2}\)

Answer \(\dfrac{(3x-1)^{6}}{18}+c\) and \(\tfrac12\sqrt{4x+1}+c\)

Example Find \(\displaystyle\int\frac{4}{2x+5}\,dx\).

Power −1: a log. Divide by a = 2.

  1. \(4\times\tfrac12\ln(2x+5)\)

Answer \(2\ln(2x+5)+c\)

Try it yourself

1
Find \(\int(2x+3)^{4}\,dx\).
Show answer

Answer \(\tfrac{1}{10}(2x+3)^{5}+c\)

2
Find \(\displaystyle\int\frac{6}{3x-2}\,dx\). (Type ln(3x-2).)
Show answer

Answer \(2\ln(3x-2)+c\)

Ways they test it

Type 1 Sum of powers

"Find \(\int(\ldots)\,dx\)." (2-3 marks)

Spot it roots, fractions or brackets of x.

  1. Rewrite as powers.
  2. Integrate term by term, + c.

Type 2 Rewrite first

"Find \(\int\dfrac{(x^{4}+1)^{2}}{2x}\,dx\)." (3 marks)

Spot it a bracket squared over a single term.

  1. Expand, divide each term.
  2. Integrate; 1/x gives ln.

Type 3 Bracket to a power

"Find \(\int\dfrac{1}{\sqrt{3x+2}}\,dx\)." (2 marks)

Spot it (ax + b) to any power.

  1. Power +1, ÷ new power ÷ a.

For example \((3x+2)^{-1/2}\)

  1. \(\dfrac{(3x+2)^{1/2}}{\frac12\times3}\)

Answer \(\tfrac23(3x+2)^{1/2}+c\)

Type 4 Log answers

"Find \(\int\dfrac{1}{4x-3}\,dx\)." (1-2 marks)

Spot it a linear bracket on the bottom, power 1.

  1. \(\tfrac1a\ln(ax+b)\).

From the past papers

Past paper 1

4037/21/M/J/23 Q7Paper 2b6 marksFull size
Find the integral of 1/(2x-6)^3 + cos x, and of (x^4+1)^2/(2x).

How to start: (i) write \(\tfrac{1}{(2x-6)^{3}}\) as \((2x-6)^{-3}\). (ii) expand the top and divide every term by 2x before integrating.

✓ Show the solution have a go on paper first
  1. (i) \(\int(2x-6)^{-3}\,dx=\dfrac{(2x-6)^{-2}}{-2\times2}=-\dfrac{1}{4(2x-6)^{2}}\); \(\int\cos x\,dx=\sin x\).
  2. (i) answer: \(-\dfrac{1}{4(2x-6)^{2}}+\sin x+c\).
  3. (ii) \(\dfrac{(x^{4}+1)^{2}}{2x}=\dfrac{x^{8}+2x^{4}+1}{2x}=\tfrac12x^{7}+x^{3}+\dfrac{1}{2x}\).
  4. (ii) answer: \(\dfrac{x^{8}}{16}+\dfrac{x^{4}}{4}+\tfrac12\ln x+c\).
PartAnswer / workingMarksGuidance
(b)(i)\(\sin x\)B1
\(-\dfrac{1}{4(2x-6)^{2}}\) (+ c)B2B1 for \(k(2x-6)^{-2}\)
(b)(ii)\(\tfrac12x^{7}+x^{3}+\tfrac{1}{2x}\)B1
\(\tfrac{x^{8}}{16}+\tfrac{x^{4}}{4}+\tfrac12\ln x+c\)B2B1 for any two terms correct
Total: 6 marks

Examiner tip Most lost marks: dividing by −2 but forgetting the 2 inside the bracket, and integrating a fraction without splitting it first.

Past paper 2

4037/23/O/N/24 Q6Paper 2a2 marksFull size
Find the integral of 1 over the square root of (3x+2).

How to start: Write it as \((3x+2)^{-1/2}\): it is a bracket to a power.

✓ Show the solution have a go on paper first
  1. Power \(-\tfrac12\to\tfrac12\); divide by \(\tfrac12\) and by 3.
  2. \(\dfrac{(3x+2)^{1/2}}{\tfrac32}=\tfrac23(3x+2)^{1/2}+c\).
PartAnswer / workingMarksGuidance
(a)\(\tfrac23(3x+2)^{1/2}+c\)B2B1 for \(k(3x+2)^{1/2}\), \(k\ne0\)
Total: 2 marks

Mistakes to avoid

  • ✗ No + c✓ every indefinite integral ends with + cIt costs the mark.
  • ✗ \(\int\frac1x\,dx=\frac{x^{0}}{0}\)✓ \(\ln x\)The power rule can't do n = −1.
  • ✗ \(\int(3x-1)^{5}dx=\frac{(3x-1)^{6}}{6}\)✓ divide by 3 as well: \(\frac{(3x-1)^{6}}{18}\)
  • ✗ \(\int\frac{x^{2}+3}{\sqrt x}dx\) done top and bottom separately✓ split into separate powers firstYou can't integrate a fraction piece by piece.

Summary

  • Integration reverses differentiation; add \(+c\).
  • \(x^{n}\to\frac{x^{n+1}}{n+1}\) (n ≠ −1); \(\frac1x\to\ln x\).
  • \((ax+b)^{n}\to\frac{(ax+b)^{n+1}}{a(n+1)}\); \(\frac{1}{ax+b}\to\frac1a\ln(ax+b)\).
  • Rewrite roots, fractions and brackets as powers before integrating.

If you see …

  • a root or a fractionrewrite as powers.
  • (ax + b) to a powerkeep the bracket, divide by a too.
  • power −1ln.

Practice

Practice: easy, medium, hard

Easy

1
\(\int 8x^{3}\,dx\)
Show answer

Answer \(2x^{4}+c\)

2
\(\displaystyle\int\frac{8}{x^{2}}\,dx\)
Show answer

Answer \(-\dfrac8x+c\)

3
\(\displaystyle\int\frac{2}{\sqrt x}\,dx\)
Show answer

Answer \(4\sqrt x+c\)

Medium

1
\(\int(x+2)^{2}\,dx\) (expand first)
Guide me

What is it asking? an integral

What do we know? \(x^{2}+4x+4\)

Plan term by term

Show working
  1. \(\tfrac{x^3}{3}+2x^2+4x+c\)

Answer \(\tfrac13x^{3}+2x^{2}+4x+c\)

2
\(\displaystyle\int\frac{3x^{2}-2x+5}{x}\,dx\)
Guide me

What is it asking? an integral

What do we know? split: \(3x-2+\tfrac5x\)

Plan integrate each

Show working
  1. \(\tfrac32x^2-2x+5\ln x+c\)

Answer \(\tfrac32x^{2}-2x+5\ln x+c\)

3
\(\int(5-2x)^{3}\,dx\)
Guide me

What is it asking? an integral

What do we know? power 4

Plan divide by 4 and by −2

Show working
  1. \(-\tfrac{(5-2x)^4}{8}+c\)

Answer \(-\tfrac18(5-2x)^{4}+c\)

Hard

1
\(\displaystyle\int\frac{1}{(2x+1)^{2}}\,dx\)
Guide me

What is it asking? an integral

What do we know? \((2x+1)^{-2}\)

Plan \(\frac{(2x+1)^{-1}}{-1\times2}\)

Show working
  1. \(-\tfrac{1}{2(2x+1)}+c\)

Answer \(-\dfrac{1}{2(2x+1)}+c\)

PartAnswer / workingMarksGuidance
\(k(2x+1)^{-1}\)B1
\(-\tfrac12(2x+1)^{-1}+c\)B1
Total: 2 marks
2
\(\displaystyle\int\frac{6}{3x-2}\,dx\)
Guide me

What is it asking? an integral

What do we know? power −1: ln

Plan \(6\times\tfrac13\)

Show working
  1. \(2\ln(3x-2)+c\)

Answer \(2\ln(3x-2)+c\)

PartAnswer / workingMarksGuidance
\(k\ln(3x-2)\)B1
\(2\ln(3x-2)+c\)B1
Total: 2 marks
3
\(\displaystyle\int\frac{x^{2}+3}{\sqrt x}\,dx\)
Guide me

What is it asking? an integral

What do we know? \(x^{3/2}+3x^{-1/2}\)

Plan powers 5/2 and 1/2

Show working
  1. \(\tfrac25x^{5/2}+6\sqrt x+c\)

Answer \(\tfrac25x^{5/2}+6x^{1/2}+c\)

PartAnswer / workingMarksGuidance
split correctlyM1
each termA1A1
Total: 3 marks

Can I do this?

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Quiz

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