4.2: Logs and Integrals

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Oct 27, 2024
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- 4.1: Logs and Derivatives
- 4.3: Exponentials With Other Bases

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Recall that

$\int \dfrac{1}{x} dx = \ln |x| + C. \nonumber$

Note that we have the absolute value sign since for negative values of that graph of $\frac{1}{x}$ is still continuous.

Example 1

Solve $\int \dfrac{dx}{1-3x}. \nonumber$

Solution

Let $u = 1-3x$ and $du = -3\, dx$ .

The integral becomes

$\begin{align*} -\dfrac{1}{3} \int \dfrac{du}{u} &= \dfrac{1}{3}\ln |u| +C \\ &= -\dfrac{1}{3} \ln |1-3x| +C. \end{align*}$

Exercise $\PageIndex{1}$

Integrate $\csc x$ .

hint: Use the formula $\csc x = \sec (\pi/2 - x)$ .

Contributors and Attributions

Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.

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Example 1

Exercise 4.2.1\PageIndex{1}

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Exercise $\PageIndex{1}$