Integrals Involving Logarithmic Functions

Integrating functions of the form \(f \left(\right. x \left.\right) = x^{−1}\) result in the absolute value of the natural log function, as shown in the following rule. Integral formulas for other logarithmic functions, such as \(f \left(\right. x \left.\right) = \text{ln} x\) and \(f \left(\right. x \left.\right) = \text{log}_{a} x ,\) are also included in the rule.

Rule: Integration Formulas Involving Logarithmic Functions

The following formulas can be used to evaluate integrals involving logarithmic functions.

\[\begin{aligned} \int x^{−1} d x & = & \text{ln} \left|\right. x \left|\right. + C \\ \int \text{ln} x d x & = & x \text{ln} x - x + C = x \left(\right. \text{ln} x - 1 \left.\right) + C \\ \int \text{log}_{a} x d x & = & \frac{x}{\text{ln} a} \left(\right. \text{ln} x - 1 \left.\right) + C \end{aligned}\]

Example 5.45

Finding an Antiderivative Involving \(\text{ln} x\)

Find the antiderivative of the function \(\frac{3}{x - 10} .\)

Solution

First factor the 3 outside the integral symbol. Then use the u⁻¹ rule. Thus,

\[\begin{aligned} \int \frac{3}{x - 10} d x & = 3 \int \frac{1}{x - 10} d x \\ \\ \\ & = 3 \int \frac{d u}{u} \\ & = 3 \text{ln} \left|\right. u \left|\right. + C \\ & = 3 \text{ln} \left|\right. x - 10 \left|\right. + C , x \neq 10 . \end{aligned}\]

See Figure 5.39.

A graph of the function f(x) = 3 / (x – 10). There is an asymptote at x=10. The first segment is a decreasing concave down curve that approaches 0 as x goes to negative infinity and approaches negative infinity as x goes to 10. The second segment is a decreasing concave up curve that approaches infinity as x goes to 10 and approaches 0 as x approaches infinity.

Figure 5.39 The domain of this function is \(x \neq 10 .\)

Checkpoint 5.38

Find the antiderivative of \(\frac{1}{x + 2} .\)

Example 5.46

Finding an Antiderivative of a Rational Function

Find the antiderivative of \(\frac{2 x^{3} + 3 x}{x^{4} + 3 x^{2}} .\)

Solution

This can be rewritten as \(\int \left(\right. 2 x^{3} + 3 x \left.\right) \left(\right. x^{4} + 3 x^{2} \left.\right)^{−1} d x .\) Use substitution. Let \(u = x^{4} + 3 x^{2} ,\) then \(d u = 4 x^{3} + 6 x .\) Alter du by factoring out the 2. Thus,

\[\begin{aligned} \\ d u & = & \left(\right. 4 x^{3} + 6 x \left.\right) d x \\ & = & 2 \left(\right. 2 x^{3} + 3 x \left.\right) d x \\ \frac{1}{2} d u & = & \left(\right. 2 x^{3} + 3 x \left.\right) d x . \end{aligned}\]

Rewrite the integrand in u:

\[\int \left(\right. 2 x^{3} + 3 x \left.\right) \left(\right. x^{4} + 3 x^{2} \left.\right)^{−1} d x = \frac{1}{2} \int u^{−1} d u .\]

Then we have

\[\begin{aligned} \frac{1}{2} \int u^{−1} d u & = \frac{1}{2} \text{ln} \left|\right. u \left|\right. + C \\ \\ & = \frac{1}{2} \text{ln} \left|\right. x^{4} + 3 x^{2} \left|\right. + C . \end{aligned}\]

Example 5.47

Finding an Antiderivative of a Logarithmic Function

Find the antiderivative of the log function \(\text{log}_{2} x .\)

Solution

Follow the format in the formula listed in the rule on integration formulas involving logarithmic functions. Based on this format, we have

\[\int \text{log}_{2} x d x = \frac{x}{\text{ln} 2} \left(\right. \text{ln} x - 1 \left.\right) + C .\]

Checkpoint 5.39

Find the antiderivative of \(\text{log}_{3} x .\)

Example 5.48 is a definite integral of a trigonometric function. With trigonometric functions, we often have to apply a trigonometric property or an identity before we can move forward. Finding the right form of the integrand is usually the key to a smooth integration.

Example 5.48

Evaluating a Definite Integral

Find the definite integral of \(\int_{0}^{\pi / 2} \frac{\text{sin} x}{1 + \text{cos} x} d x .\)

Solution

We need substitution to evaluate this problem. Let \(u = 1 + \text{cos} x , ,\) so \(d u = − \text{sin} x d x .\) Rewrite the integral in terms of u, changing the limits of integration as well. Thus,

\[\begin{aligned} u = 1 + \text{cos} \left(\right. 0 \left.\right) = 2 \\ u = 1 + \text{cos} \left(\right. \frac{\pi}{2} \left.\right) = 1 . \end{aligned}\]

Then

\[\begin{aligned} \int_{0}^{\pi / 2} \frac{\text{sin} x}{1 + \text{cos} x} & = − \int_{2}^{1} u^{−1} d u \\ \\ \\ & = \int_{1}^{2} u^{−1} d u \\ & = \left(\text{ln} \left|\right. u \left|\right. \left|\right.\right)_{1}^{2} \\ & = \left[\right. \text{ln} 2 - \text{ln} 1 \left]\right. \\ & = \text{ln} 2. \end{aligned}\]

This lesson is part of:

Integration

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