1.10.1.2: Logs and Integrals

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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 \(\csc x = \sec (\pi/2 - x)\).

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