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Integration of reciprocal functions

How do you integrate a reciprocal function? This skill is important for calculating electrical currents in circuits and modelling harmonic motion in mechanical systems. Use this resource to learn how.

Reciprocal functions have the form:
\[y=\frac{k}{x}\]
where \(k\) is a constant. We often deal with reciprocal functions when looking at Inverse functions. Examples include \(y=\dfrac{1}{x}\) and \(f(x)=\dfrac{1}{3x-1}\).

The power rule

One of the most important rules for integration is the power rule. It states that:

\[ \int x^{n}dx=\frac{1}{n+1}x^{n+1}+c,\,n\neq-1 \]

where \(c\) is a constant.

The reason that \(n\neq-1\) is that if \(n=-1\), \(\dfrac{1}{n+1}\) would be \(\dfrac{1}{0}\), which is undefined. \(n\) can be any number (other than \(-1\)) — positive, negative or a fraction. Often, instead of saying \(c\) is a constant, we write \(c\in\mathbb{R}\) which means \(c\) is a real number.

When we encounter a case when \(n=-1\), such as integrating \(\int\left(\dfrac{1}{x}\right)dx\), we define a new function:

\[ \int\frac{1}{x}dx=\log_{e}\left|x\right|+c\]

where \(c\) is a constant.

The function \(\log_{e}\left|x\right|\) is called the natural logarithm. It is sometimes written as \(\ln\left|x\right|\). The absolute value sign is used as the natural logarithm is not defined for negative arguments.

We can now integrate \(f(x)=\dfrac{m}{x}\) where \(m\) is any constant:

\[\begin{align*} \int\frac{m}{x}dx & = m\int\frac{1}{x}dx\\
& = m\log_{e}\left|x\right|+c \end{align*}\]

There is a more general rule:

\[ \int\frac{m}{ax+b}dx=\frac{m}{a}\log_{e}\left|ax+b\right|+c \]

where \(a\), \(b\) and \(m\) be constants with \(a\neq0\).

Example 1 – integrating reciprocal functions

Integrate \(f(x)=\dfrac{1}{2x}\) with respect to \(x\).

Here, \(m=1\), \(a=2\) and \(b=0\).
\[\int\frac{1}{2x}dx=\frac{1}{2}\log_{e}\left|2x\right|+c,\, c\in\mathbb{R}\]

Calculate the integral \(\int\dfrac{2}{3x}dx\).

Here, \(m=2\), \(a=3\) and \(b=0\).
\[\int\frac{2}{3x}dx=\frac{2}{3}\log_{e}\left|3x\right|+c,\,c\in\mathbb{R}\]

Integrate \(f(x)=\dfrac{13}{5x-7}\) with respect to \(x\).

Here, \(m=13\), \(a=5\) and \(b=-7\).
\[\int\frac{13}{5x-7}dx=\frac{13}{5}\ln\left|5x-7\right|+c,\,c\in\mathbb{R}\]

Integrate \(f(x)=\dfrac{6}{2-3x}\) with respect to \(x\).

Here, \(m=6\), \(a=-3\) and \(b=2\).
\[\begin{align*} \int\frac{6}{2-3x}dx & = \frac{6}{-3}\log_{e}\left|2-3x\right|+c,\,c\in\mathbb{R}\\
& = -2\log_{e}\left|2-3x\right|+c
\end{align*}\]

Integrate \(f(x)=\dfrac{6}{2-3x}-\dfrac{2}{x+1}\) with respect to \(x\).

Here, \(m=6\), \(a=-3\) and \(b=2\) in the first term on the right-hand side and \(m=-2\), \(a=1\) and \(b=1\) in the second.

\[\begin{align*} \int\left(\frac{6}{2-3x}-\frac{2}{x+1}\right)dx & = \frac{6}{-3}\log_{e}\left|2-3x\right|-2\log_{e}\left|x+1\right|+c,\,c\in\mathbb{R}\\
& = -2\log_{e}\left|2-3x\right|-2\log_{e}\left|x+1\right|+c \end{align*}\]

Find the integral \(\int_{1}^{2}\dfrac{1}{2x}dx\).

In this case, the antiderivative is \(F(x)=\dfrac{1}{2}\log_{e}\left|2x\right|+c,\,c\in\mathbb{R}\).

It is also handy to remember the log laws.
\[\begin{align*} \int_{1}^{2}\frac{1}{2x}dx & = \left[\frac{1}{2}\log_{e}\left|2x\right|+c\right]_{x=1}^{x=2}\\
& = \frac{1}{2}\log_{e}\left|2(2)\right|+c-\left(\frac{1}{2}\log_{e}\left|2\left(1\right)\right|+c\right)\\
& = \frac{1}{2}\log_{e}\left|4\right|-\frac{1}{2}\log_{e}\left|2\right|\\
& = \frac{1}{2}\log_{e}\left|\frac{4}{2}\right|\\
& = \frac{1}{2}\log_{e}(2)
\end{align*}\]

Find the integral \(\int_{2}^{3}\dfrac{13}{5x-7}dx\).

In this case, the antiderivative is \(F(x)=\dfrac{13}{5}\log_{e}\left|5x-7\right|+c,\,c\in\mathbb{R}\).
\[\begin{align*} \int_{2}^{3}\frac{13}{5x-7}dx & = \left[\frac{13}{5}\log_{e}\left|5x-7\right|\right]_{2}^{3}\\
& = \frac{13}{5}\log_{e}\left|15-7\right|-\left(\frac{13}{5}\log_{e}\left|10-7\right|\right)\\
& = \frac{13}{5}\log_{e}\left|8\right|-\left(\frac{13}{5}\log_{e}\left|3\right|\right)\\
& = \frac{13}{5}\log_{e}\frac{8}{3}
\end{align*}\]

Exercise – integrating reciprocal functions

  1. Calculate the following integrals.
    1. \(\int\dfrac{1}{5x}dx\)
    2. \(\int\dfrac{3}{2x}dx\)
    3. \(\int\dfrac{3}{3x-5}dx\)
    4. \(\int\dfrac{6}{2-7x}dx\)
  2. Integrate the following functions with respect to \(x\).
    1. \(f(x)=x^{2}-\dfrac{2}{x}\)
    2. \(f(x)=\dfrac{1}{3x-2}+\dfrac{3}{1-x}\)
    3. \(f(x)=\dfrac{3}{3x-7}-x^{2}\)
    4. \(f(x)=\dfrac{2}{5x-4}-\dfrac{3}{2-5x}\)
  3. Find the following intervals.
    1. \(\int_{1}^{5}\dfrac{2}{3x}dx\)
    2. \(\int_{1}^{2}\dfrac{2}{3x-1}dx\)
    3. \(\int_{1}^{3}\left(x^{2}-\dfrac{2}{x}\right)dx\)
    4. \(\int_{2}^{4}\left(\dfrac{2}{5x-4}-\dfrac{3}{2-5x}\right)dx\)

Hint: \(\log_{e}(1)=0\).

    1. \(\dfrac{1}{5}\log_{e}\left|x\right|+c\)
    2. \(\dfrac{3}{2}\log_{e}\left|2x\right|+c\)
    3. \(\log_{e}\left|3x-5\right|+c\)
    4. \(-\dfrac{6}{7}\log_{e}\left|2-7x\right|+c\)
    1. \(\dfrac{1}{3}x^{3}-2\log_{e}\left|x\right|+c\)
    2. \(\dfrac{1}{3}\log_{e}\left|3x-2\right|-3\log_{e}\left|1-x\right|+c\)
    3. \(\log_{e}\left|3x-5\right|-\dfrac{1}{3}x^{3}+c\)
    4. \(\dfrac{2}{5}\log_{e}\left|5x-4\right|+\dfrac{3}{5}\log_{e}\left|2-5x\right|+c\)
    1. \(\dfrac{2}{3}\log_{e}(5)\)
    2. \(\dfrac{2}{3}\log_{e}\left(\dfrac{5}{2}\right)\)
    3. \(\dfrac{26}{3}-2\log_{e}(3)\)
    4. \(\dfrac{2}{5}\log_{e}\left(\dfrac{8}{3}\right)+\dfrac{3}{5}\log_{e}\left(\dfrac{9}{4}\right)\)

Integration of reciprocal functions in context

In business, the rate at which goods are transported across a network is modelled by the function \(y=\dfrac{5}{x}\), where \(x\) represents the distance travelled. To calculate the total volume of goods moved over varying distances, integrate this function with respect to \(x\).

\[\begin{align*} \int\left(\frac{5}{x}\right)dx & = 5\int\left(\frac{1}{x}\right)dx\\
& = 5\ln(x)+c
\end{align*}\]

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