Standard Integrals
IB Mathematics AA foundations: core antiderivatives, linearity and linear inputs.
On this page
- Core standard integrals
- Linearity of integration
- Worked examples: standard integrals
- Linear inputs of the form \(ax+b\)
- Worked examples: linear inputs
- Common traps
- Practice questions
- Answer key
1Core standard integrals
The following are the standard results students should be able to use fluently. For indefinite integrals, always include the constant of integration \(C\).
| Integral | Result and notes |
|---|---|
| \(\displaystyle \int k\,dx\) | \(kx+C\), where \(k\) is a constant. |
| \(\displaystyle \int x^n\,dx\) | \(\displaystyle \frac{x^{n+1}}{n+1}+C\), for \(n\in\mathbb{Q}\) and \(n\ne -1\). |
| \(\displaystyle \int \frac{1}{x}\,dx\) | \(\ln|x|+C\). If the context restricts \(x>0\), this may be written as \(\ln x+C\). |
| \(\displaystyle \int e^x\,dx\) | \(e^x+C\). |
| \(\displaystyle \int \sin x\,dx\) | \(-\cos x+C\). |
| \(\displaystyle \int \cos x\,dx\) | \(\sin x+C\). |
The power rule does not work for \(n=-1\), because it would require division by zero. That is why
\[ \int x^{-1}\,dx=\int \frac{1}{x}\,dx=\ln|x|+C. \]2Linearity: combining standard integrals
Before moving to inputs of the form \(ax+b\), students should be confident using the standard table with sums, differences and constant multiples.
where \(a\) and \(b\) are constants. In words: integrate term by term, keeping constant multipliers in front.
Standard results
\[ \int f(x)\,dx=F(x)+C \] \[ \int g(x)\,dx=G(x)+C \]Linearity
\[ \int \big(af(x)+bg(x)\big)\,dx=aF(x)+bG(x)+C \]Before integrating, rewrite roots and fractions as powers when this makes the standard table easier to use. For example, \(\sqrt{x}=x^{1/2}\) and \(\frac{1}{\sqrt{x}}=x^{-1/2}\).
3Worked examples: using linearity with standard integrals
Worked example 1
Find \(\displaystyle \int \left(6x^2-\frac{4}{x}+5e^x\right)\,dx\).
Worked example 2
Find \(\displaystyle \int (3\sin x-2\cos x)\,dx\).
Worked example 3
Find \(\displaystyle \int \left(4x^3-3\sqrt{x}+\frac{2}{\sqrt{x}}\right)\,dx\).
First write \(\sqrt{x}=x^{1/2}\) and \(\frac{1}{\sqrt{x}}=x^{-1/2}\).
\[ \begin{aligned} \int \left(4x^3-3x^{1/2}+2x^{-1/2}\right)\,dx &=x^4-3\cdot \frac{x^{3/2}}{3/2}+2\cdot \frac{x^{1/2}}{1/2}+C\\ &=x^4-2x^{3/2}+4x^{1/2}+C. \end{aligned} \]4Linear inputs: functions of \(ax+b\)
Once the standard results and linearity are secure, we can change the input from \(x\) to a linear expression \(ax+b\).
Standard result
\[ \int f(x)\,dx=F(x)+C \]Linear input
\[ \int f(ax+b)\,dx=\frac{1}{a}F(ax+b)+C \]If \(F'(x)=f(x)\) and \(a\ne0\), then
\[ \int f(ax+b)\,dx=\frac{1}{a}F(ax+b)+C. \]In words: integrate as usual, then divide by the coefficient of \(x\) inside the bracket.
When the inside is \(ax+b\), its derivative is \(a\). So the antiderivative must be adjusted by a factor of \(\frac{1}{a}\). Do not divide by \(b\).
| Integral | Result |
|---|---|
| \(\displaystyle \int (ax+b)^n\,dx\) | \(\displaystyle \frac{(ax+b)^{n+1}}{a(n+1)}+C\), for \(n\ne -1\). |
| \(\displaystyle \int \frac{1}{ax+b}\,dx\) | \(\displaystyle \frac{1}{a}\ln|ax+b|+C\). |
| \(\displaystyle \int e^{ax+b}\,dx\) | \(\displaystyle \frac{1}{a}e^{ax+b}+C\). |
| \(\displaystyle \int \sin(ax+b)\,dx\) | \(\displaystyle -\frac{1}{a}\cos(ax+b)+C\). |
| \(\displaystyle \int \cos(ax+b)\,dx\) | \(\displaystyle \frac{1}{a}\sin(ax+b)+C\). |
This page deliberately stops at standard results, linearity and linear composites. Reverse chain rule, substitution and more advanced HL integration techniques will be treated on separate pages.
5Worked examples: linear inputs
Worked example 4
Find \(\displaystyle \int \cos(2x+3)\,dx\).
Since \(\int \cos u\,du=\sin u+C\), and the coefficient of \(x\) inside is \(2\),
\[ \int \cos(2x+3)\,dx=\frac{1}{2}\sin(2x+3)+C. \]Worked example 5
Find \(\displaystyle \int \frac{1}{5-2x}\,dx\).
Here \(5-2x=-2x+5\), so \(a=-2\).
\[ \int \frac{1}{5-2x}\,dx=-\frac{1}{2}\ln|5-2x|+C. \]Check
\[ \frac{d}{dx}\left(-\frac{1}{2}\ln|5-2x|\right)=-\frac{1}{2}\cdot\frac{-2}{5-2x}=\frac{1}{5-2x}. \]Worked example 6
Find \(\displaystyle \int \left(4x^3-3\sqrt{x}+2e^{1-3x}\right)\,dx\).
First write \(\sqrt{x}=x^{1/2}\). Then integrate term by term.
\[ \begin{aligned} \int \left(4x^3-3x^{1/2}+2e^{1-3x}\right)\,dx &=x^4-3\cdot\frac{x^{3/2}}{3/2}+2\left(-\frac{1}{3}e^{1-3x}\right)+C\\ &=x^4-2x^{3/2}-\frac{2}{3}e^{1-3x}+C. \end{aligned} \]Worked example 7
Find \(\displaystyle \int (3x-5)^7\,dx\).
The inside is \(3x-5\), so \(a=3\).
\[ \int (3x-5)^7\,dx=\frac{(3x-5)^8}{3\cdot8}+C=\frac{(3x-5)^8}{24}+C. \]Check Differentiating \(\frac{(3x-5)^8}{24}\) gives
\[ \frac{1}{24}\cdot 8(3x-5)^7\cdot 3=(3x-5)^7. \]6Common traps
| Trap | Fix |
|---|---|
| Forgetting \(+C\) | Include \(+C\) for indefinite integrals, unless an initial condition is used to find a particular antiderivative. |
| Using the power rule for \(x^{-1}\) | Use \(\displaystyle \int \frac{1}{x}\,dx=\ln|x|+C\). |
| Forgetting the linear coefficient | For \(f(ax+b)\), divide by \(a\). |
| Dividing by the wrong number | Divide by the coefficient of \(x\), not the constant term. |
| Mixing degrees and radians | In calculus, trigonometric functions are interpreted in radians unless explicitly stated otherwise. |
| Dropping absolute values in logarithms | Write \(\ln|ax+b|\) unless the domain guarantees \(ax+b>0\). |
7Practice
Try these without looking at the answers first. For indefinite integrals, include \(+C\).
A. Core standard integrals and linearity
- A1\(\displaystyle \int 7x^6\,dx\)
- A2\(\displaystyle \int (5x^4-3x^2+8)\,dx\)
- A3\(\displaystyle \int \left(\frac{6}{x}+4e^x\right)\,dx\)
- A4\(\displaystyle \int (3\sin x-2\cos x)\,dx\)
- A5\(\displaystyle \int \left(\sqrt{x}+\frac{2}{\sqrt{x}}\right)\,dx\)
B. Linear inputs
- B1\(\displaystyle \int (2x+1)^5\,dx\)
- B2\(\displaystyle \int (7-3x)^4\,dx\)
- B3\(\displaystyle \int e^{4x-1}\,dx\)
- B4\(\displaystyle \int \sin(5x)\,dx\)
- B5\(\displaystyle \int \cos(3x-\pi)\,dx\)
- B6\(\displaystyle \int \frac{1}{4x+9}\,dx\)
- B7\(\displaystyle \int \frac{3}{2-5x}\,dx\)
C. Mixed and particular antiderivatives
- C1\(\displaystyle \int \big(2(3x-1)^4-5e^{-x}\big)\,dx\)
- C2\(\displaystyle \int \left(6x^2+4\cos(2x)-\frac{3}{x}\right)\,dx\)
- C3\(\displaystyle \int \big(5e^{2x-1}-2\sin(3x)\big)\,dx\)
- C4Find \(F(x)\) if \(F'(x)=6x^2-\frac{2}{x}\) and \(F(1)=4\).
- C5Explain why \(\displaystyle \int x^{-1}\,dx\) is not found using \(\displaystyle \frac{x^0}{0}\).
8Answer key
Show answers
Answer key labelled to match the practice sections: A1–A5, B1–B7 and C1–C5.
A. Core standard integrals and linearity
- A1\(x^7+C\)
- A2\(x^5-x^3+8x+C\)
- A3\(6\ln|x|+4e^x+C\)
- A4\(-3\cos x-2\sin x+C\)
- A5\(\displaystyle \frac{2}{3}x^{3/2}+4x^{1/2}+C\)
B. Linear inputs
- B1\(\displaystyle \frac{(2x+1)^6}{12}+C\)
- B2\(\displaystyle -\frac{(7-3x)^5}{15}+C\)
- B3\(\displaystyle \frac{1}{4}e^{4x-1}+C\)
- B4\(\displaystyle -\frac{1}{5}\cos(5x)+C\)
- B5\(\displaystyle \frac{1}{3}\sin(3x-\pi)+C\)
- B6\(\displaystyle \frac{1}{4}\ln|4x+9|+C\)
- B7\(\displaystyle -\frac{3}{5}\ln|2-5x|+C\)
C. Mixed and particular antiderivatives
- C1\(\displaystyle \frac{2(3x-1)^5}{15}+5e^{-x}+C\)
- C2\(2x^3+2\sin(2x)-3\ln|x|+C\)
- C3\(\displaystyle \frac{5}{2}e^{2x-1}+\frac{2}{3}\cos(3x)+C\)
- C4\(F(x)=2x^3-2\ln|x|+2\)
- C5The power rule requires division by \(n+1\). If \(n=-1\), then \(n+1=0\), so the rule is not valid. Instead, \(\displaystyle \int x^{-1}\,dx=\ln|x|+C\).