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Antiderivatives of exponential and logarithmic functions

We've seen various derivatives so far, including

\[ \dfrac{d}{dx}\, e^x = e^x \qquad \text{and, more generally,} \qquad \dfrac{d}{dx}\, e^{kx} = k \, e^{kx}, \] where \(k\) is any non-zero real constant. Also, we've seen \[ \dfrac{d}{dx}\, \log_e x = \dfrac{1}{x} \qquad \text{and, more generally,} \qquad \dfrac{d}{dx}\, \log_e(ax+b) = \dfrac{a}{ax+b}, \] for \(ax+b>0\), where \(a,b\) are real constants with \(a \ne 0\). From this we can deduce several antiderivatives.

The basic indefinite integrals are

\[ \int e^x \; dx = e^x + c \qquad \text{and} \qquad \int \dfrac{1}{x} \; dx = \log_e x + c \] and, more generally, \[ \int e^{kx} \; dx = \dfrac{1}{k} \, e^{kx} + c \qquad \text{and} \qquad \int \dfrac{1}{ax+b} \; dx = \dfrac{1}{a} \, \log_e(ax+b) + c, \] where \(c\) as usual is a constant of integration.

We can use these antiderivatives to evaluate definite integrals.

Example

Find

\[ \int_e^{e^2} \dfrac{1}{x} \; dx. \]

Solution

\begin{align*} \int_e^{e^2} \dfrac{1}{x} \; dx &= \bigl[ \log_e x \bigr]_e^{e^2}\\ &= \log_e (e^2) - \log_e e = 2-1 = 1 \end{align*}

Exercise 11

Prove that, for any \(x>0\),

\[ \int_1^x \dfrac{1}{t} \; dt = \log_e x. \]

Exercise 12

Prove that, for any \(x>0\) and \(n>0\),

\[ \int_{x^n}^{x^{n+1}} \dfrac{1}{t} \; dt = \log_e x. \]

Exercise 13

Differentiate \(f(x) = x \log_e x - x\). Hence find the indefinite integral

\[ \int \log_e x \; dx. \]

Screencast of exercise 13

Warning! As we mentioned previously, \(\log_e x\) is only defined for \(x>0\), while \(\dfrac{1}{x}\) is defined for all \(x \ne 0\). So the equation \[ \int \dfrac{1}{x} \; dx = \log_e x + c \] is valid only for \(x>0\) and, more generally, the equation \[ \int \dfrac{1}{ax+b} \; dx = \dfrac{1}{a} \, \log_e (ax+b) + c \] is valid only when \(ax+b>0\). Although it is more complicated, it is sometimes necessary to consider the function \(\log_e|x|\), which is defined for all \(x \neq 0\) and which also has derivative \(\dfrac{1}{x}\). The equation \[ \int \dfrac{1}{x} \; dx = \log_e |x| + c \] is valid for all \(x \neq 0\), and the equation \[ \int \dfrac{1}{ax+b} \; dx = \dfrac{1}{a} \log_e |ax+b| + c \] is valid for all \(x\) such that \(ax+b \neq 0\).

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