This note covers anti-differentiation: finding a function whose derivative you are given. You will learn the standard integrals, how to handle linear inside functions, and how to spot a "function and its derivative" pattern using the reverse chain rule or a substitution. It is the toolkit behind areas, kinematics and many modelling questions.
Key ideas
Integration is the reverse of differentiation. If \(F'(x)=f(x)\), then \(\int f(x)\,dx=F(x)+C\). The constant \(C\) is needed because every constant disappears when we differentiate. If you are given a point on the curve, use it to find \(C\).
Basic results. Rewrite roots and reciprocals as powers first: \(\sqrt{x}=x^{1/2}\) and \(\dfrac1{x^3}=x^{-3}\). Then raise the power by one and divide by the new power. The only power that does not work is \(x^{-1}\), whose integral is \(\ln|x|\).
Integrals can be split term by term, and constant multiples come outside: \(\int (af(x)+bg(x))\,dx=a\int f\,dx+b\int g\,dx\). There is no similar rule for products or quotients. You cannot integrate \(x\,e^{x}\) by integrating each factor separately.
Linear inside function. If \(\int f(x)\,dx=F(x)+C\), then \[\int f(ax+b)\,dx=\frac1a F(ax+b)+C.\] For example, \(\int \cos(3x-2)\,dx=\tfrac13\sin(3x-2)+C\). Check by differentiating: the chain rule brings back the factor 3, which cancels the \(\tfrac13\).
Reverse chain rule. Differentiating \(F(g(x))\) gives \(F'(g(x))\,g'(x)\). So, going backwards, \[\int f'(g(x))\,g'(x)\,dx=f(g(x))+C.\] This applies when the integrand is a composite function multiplied by (a multiple of) the derivative of the inside function. A very common special case is \[\int \frac{g'(x)}{g(x)}\,dx=\ln|g(x)|+C.\]
Substitution. When the pattern is less obvious, write \(u=g(x)\), find \(\dfrac{du}{dx}\), replace every \(x\) and \(dx\) so that only \(u\) remains, integrate, then put \(g(x)\) back. If any \(x\) is left over after the swap, the substitution has not worked, or a different choice of \(u\) is needed. Usually \(u\) is the inside of a bracket, a power, or the denominator.
Expanding first. Products of brackets and fractions with a single-term denominator are often easiest after algebra. For example, \(\dfrac{x^3+2x}{x}=x^2+2\), which is simple to integrate, while \((x+2)^2\) can either be expanded or integrated with the linear rule to give \(\tfrac13(x+2)^3\). The two answers differ only by a constant, so both are correct.
Choosing a method. Work through these questions in order. Is it a sum of powers? Rewrite and use the power rule. Is the inside function linear? Divide by its coefficient. Is there a composite function with (a multiple of) its inside derivative next to it? Use the reverse chain rule or substitute. If none of these apply, the integral is probably not intended to be done by these methods.
Finding the constant. A question may give a gradient function and a point, or a rate of change and a starting value. Integrate first, then substitute the known values to find \(C\). Always substitute into the integrated function, not into the original gradient.
Check by differentiating. You can always verify an answer by differentiating it. The result must equal the original integrand.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C\) | For any \(n\ne -1\), including negative and fractional powers. |
| \(\int \dfrac1x\,dx=\ln|x|+C\) | The exceptional power, \(n=-1\). |
| \(\int e^x\,dx=e^x+C\) | The exponential function is its own anti-derivative. |
| \(\int \sin x\,dx=-\cos x+C\) | Note the minus sign. |
| \(\int \cos x\,dx=\sin x+C\) | Angles in radians. |
| \(\int f(ax+b)\,dx=\dfrac1aF(ax+b)+C\) | Linear inside function. Divide by the coefficient of \(x\). |
| \(\int f'(g(x))\,g'(x)\,dx=f(g(x))+C\) | Reverse chain rule: composite times derivative of the inside. |
| \(\int \dfrac{g'(x)}{g(x)}\,dx=\ln|g(x)|+C\) | Numerator is the derivative of the denominator. |
| \(u=g(x),\ du=g'(x)\,dx\) | Substitution: rewrite the whole integral in terms of \(u\). |
Worked examples
Example 1 Find \(\displaystyle\int\left(6x^2-\frac{4}{x^3}+3\sqrt{x}\right)dx\).
Rewrite with powers: \(6x^2-4x^{-3}+3x^{1/2}\).
Integrate term by term: \(6\cdot\dfrac{x^3}{3}-4\cdot\dfrac{x^{-2}}{-2}+3\cdot\dfrac{x^{3/2}}{3/2}\).
Simplify: \(2x^3+2x^{-2}+2x^{3/2}\).
Answer: \(2x^3+\dfrac{2}{x^2}+2x^{3/2}+C\).
Example 2 Find (a) \(\displaystyle\int x(x^2+5)^4\,dx\), (b) \(\displaystyle\int x\,e^{x^2}\,dx\), (c) \(\displaystyle\int \frac{\cos x}{\sin x}\,dx\).
(a) Let \(u=x^2+5\). Then \(\dfrac{du}{dx}=2x\), so \(x\,dx=\tfrac12\,du\).
The integral becomes \(\displaystyle\int u^4\cdot\tfrac12\,du=\tfrac12\cdot\frac{u^5}{5}+C=\frac{u^5}{10}+C\).
Put \(u\) back: \(\dfrac{(x^2+5)^5}{10}+C\).
(b) By inspection, the inside function is \(x^2\) and its derivative is \(2x\). Differentiating \(e^{x^2}\) gives \(2x\,e^{x^2}\), which is twice what we want. So the answer is \(\tfrac12e^{x^2}+C\).
(c) The numerator \(\cos x\) is the derivative of the denominator \(\sin x\), so the integral is \(\ln|\sin x|+C\).
Answer: (a) \(\dfrac{(x^2+5)^5}{10}+C\); (b) \(\tfrac12e^{x^2}+C\); (c) \(\ln|\sin x|+C\).
Example 3 The gradient of a curve is \(f'(x)=\dfrac{4}{2x+1}\) for \(x\gt -\tfrac12\). The curve passes through \((0,3)\). Find \(f(x)\) and then \(f(2)\).
The inside function \(2x+1\) is linear with coefficient 2, so \(\displaystyle\int\frac{4}{2x+1}\,dx=4\cdot\frac12\ln|2x+1|+C=2\ln|2x+1|+C\).
Use the point \((0,3)\): \(3=2\ln 1+C=C\), so \(C=3\).
So \(f(x)=2\ln(2x+1)+3\) (the modulus is not needed since \(2x+1\gt 0\)).
Then \(f(2)=2\ln 5+3=3.2189\ldots+3=6.22\) to 3 s.f.
Answer: \(f(x)=2\ln(2x+1)+3\), and \(f(2)\approx 6.22\).
Common errors
- Common error: leaving out \(+C\) in an indefinite integral, or forgetting to find \(C\) when a point is given.
- Common error: integrating \(\dfrac1{x^3}\) as \(\ln\) of something. Only \(\dfrac1x\) gives a logarithm. Rewrite \(\dfrac1{x^3}\) as \(x^{-3}\).
- Common error: forgetting to divide by the coefficient of \(x\) in \(\int f(ax+b)\,dx\), for example writing \(\int e^{5x}\,dx=e^{5x}+C\).
- Common error: sign slips with trigonometry. \(\int\sin x\,dx=-\cos x+C\), but \(\int\cos x\,dx=+\sin x+C\).
- Common error: leaving some \(x\)-terms in a substitution, or forgetting to replace \(u\) by the original expression at the end.
Exam tip: After finding an anti-derivative, differentiate it in your head. If you do not get back the original expression, check for a missing constant factor from the chain rule.
Check yourself
- Find \(\displaystyle\int\left(4x^3-\frac2x+5\right)dx\).
Answer
\(x^4-2\ln|x|+5x+C\). - Find \(\displaystyle\int\cos(2x+1)\,dx\).
Answer
The inside function is linear with coefficient 2, so the result is \(\tfrac12\sin(2x+1)+C\). - Find \(\displaystyle\int x^2e^{x^3}\,dx\).
Answer
Differentiating \(e^{x^3}\) gives \(3x^2e^{x^3}\). So the integral is \(\tfrac13e^{x^3}+C\). - A curve has gradient \(\dfrac{dy}{dx}=6x(x^2+1)^2\) and passes through \((0,5)\). Find \(y\) when \(x=1\).
Answer
Let \(u=x^2+1\), so \(du=2x\,dx\) and \(6x\,dx=3\,du\). Then \(y=\int3u^2\,du=u^3+C=(x^2+1)^3+C\). At \((0,5)\): \(5=1+C\), so \(C=4\). At \(x=1\): \(y=2^3+4=12\).