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Indefinite Integration

Definition and Notation

Indefinite integration can be regarded as the opposite, or inverse, of differentiation. For this reason, an indefinite integral is also referred to as an antiderivative.

The indefinite integral of a function $f(x)$ with respect to $x$ is denoted:

\[\int f(x) \mathrm{d} x.\]

The function appearing inside the integral, $f(x)$, is known as the integrand.

Given a function $f(x)=F'(x)$, where the prime denotes the derivative with respect to $x$, the indefinite integral of $f(x)$ is

\[\int f(x) \mathrm{d} x = F(x) + C,\]

$C$ is an arbitrary constant known as the constant of integration and arises from the constant rule: for a constant function $g(x)=C$, where $C$ is any number, the derivative of $g(x)$ is $g'(x)=0.$ Hence the derivative of any function $F(x)+C$ is $F'(x) = f(x)$.The value of $C$ might be determined by boundary value conditions on the function $F(x)$.

Note: Conventionally, lower-case letters are used to denote derivatives and upper-case letters are used to denote the corresponding antiderivative.

Properties

  •        The integral of a linear combination (sum) of functions is equal to a linear combination of the integrals:
           \[\int f(x)+g(x) \, \mathrm{d}x = \int f(x) \, \mathrm{d}x + \int g(x) \, \mathrm{d}x .\]

    This means that integrals can be split up into their separate terms.

  •        The integral of the product of a constant and a function is equal to the product of the constant and the integral:
           \[\int\alpha f(x) \, \mathrm{d}x = \alpha\int f(x) \, \mathrm{d}x .\]

    This means that constant factors can be taken outside the integral sign.

Table of Integrals

Here is a table of standard integrals, where $k$ is a constant.

$f(x)$

$\int{f(x)}\,\mathrm{d}x$

$k \neq 0$

$kx+c$

$x^n, \, n\neq-1$

$\dfrac{x^{n+1} }{n+1}+c$

$x^{-1}=\dfrac{1}{x}$

$\ln{x}+c$

$\mathrm{e}^{kx}, \,k\neq0$

$\dfrac{\mathrm{e}^{kx} }{k}+c$

$\sin{(kx)}, \,k\neq0$

$-\dfrac{\cos{(kx)} }{k}+c$

$\cos{(kx)}, \,k\neq0$

$\dfrac{\sin{(kx)} }{k}+c$

$\sec^2{(kx)}, \, k\neq0$

$\dfrac{\tan{(kx)} }{k}+c$

A more extensive table of integrals can be found in the handy mathcentre Facts and Formulae leaflet.

Worked Example

Example 1

Find $F(x) = \displaystyle{ \int 2x^5+3\sin{x}\;\mathrm{d}x. }$

Solution

By the above notation we have $f(x)=2x^5+3\sin{x}$, where $f(x)=F'(x)$. This demonstrates the convention of using upper-case letters to denote the antiderivative, and lower-case letters to denote the derivative (i.e. the function appearing within the integral).

By the properties of integration, the integral can be split up into two integrals and the constant factors can be pulled outside of each integral:

\[\int 2x^5+3\sin{x} \, \mathrm{d}x=2\int x^5 \, \mathrm{d}x + 3\int \sin{x} \, \mathrm{d}x\]

Each integral is computed using the standard rules.

The first integral is:

\[\int x^5 \, \mathrm{d}x = \dfrac{1}{6}x^6+\text{constant}.\]

It can be shown that $\dfrac{1}{6}x^6$ is indeed the antiderivative of $x^5$ by differentiating:

\[\dfrac{\mathrm{d} }{\mathrm{d}x}\Bigl[\dfrac{1}{6}x^6\Bigl]=\dfrac{1}{6}\cdot 6\cdot x^5=x^5.\]

The second integral is:

\[\int\sin{x} \, \mathrm{d}x = -\cos{x}+\text{constant},\]

having noted that the derivative of $\cos{x}$ is:

\[\dfrac{\mathrm{d} }{\mathrm{d}x}\bigl[\cos{x}]=-\sin{x},\]

and the derivative of $-\cos{x}$ is therefore:

\[\dfrac{\mathrm{d} }{\mathrm{d}x}\bigl[-\cos{x}]=\sin{x}.\]

Combining these results gives:

\begin{align} F(x) &= \int 2x^5+3\sin(x) \, \mathrm{d}x \\ &= 2\int x^5 \, \mathrm{d}x + 3\int \sin{x} \, \mathrm{d}x \\ &= 2\Bigl[\dfrac{1}{6}x^6\Bigl]+3\bigl[-\cos{x}\bigl]+C \\ &=\dfrac{x^6}{3}-3\cos{x}+C, \end{align}

where $C$, the sum of the two constants, is the arbitrary constant of integration.

Note: When evaluating indefinite integrals it is useful to check the result by differentiating. Here $F(x)= \dfrac{x^6}{3}-3\cos{x}+C$, which should satisfy $F'(x)=f(x)=2x^5+3\sin{x}$.

The derivative of $F(x)$ is:

\begin{align} \frac{\mathrm{d

{\mathrm{d} x}\left[ \frac{x^6}{3}-3\cos{x}+C \right] &= \frac{1}{3}\cdot6x^5-3\cdot(-\sin{x})+0 \\ &=2x^5+3\sin{x} \\ &= f(x). \end{align}

This shows that the solution is indeed correct.

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Video Examples

Example 1

Prof. Robin Johnson computes $\begin{align}\int{(x^4+2x^{-3}+1)}\;\mathrm{d}x\end{align}$ and $\begin{align}\int{\left(\mathrm{e}^{\large{2x} }+\cos{(2x)}\right)}\;\mathrm{d}x\end{align}$.

Example 2

Prof. Robin Johnson computes $\begin{align}\int\dfrac{2x-5}{x^2-5x+3}\;\mathrm{d}x\end{align}$, an integral of the form $\begin{align}\int\dfrac{f'(x)}{f(x)}\end{align}$.

Example 3

Prof. Robin Johnson computes $\begin{align}\int\dfrac{x+2}{(2x-1)^4}\;\mathrm{d}x\end{align}$.

Example 4

Prof. Robin Johnson computes $\begin{align}\int\dfrac{\bigl[\ln{|2x-3|}\bigl]^5}{2x-3}\;\mathrm{d}x\end{align}$, an integral of the form $\begin{align}\int f'(x)g(f(x))\;\mathrm{d}x\end{align}$.

Workbooks

These workbooks produced by HELM are good revision aids, containing key points for revision and many worked examples.

Test Yourself

Test yourself: Numbas test on indefinite integration

External Resources

Whiteboard maths

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