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Mathematics LibreTexts

4.2: Logs and Integrals

  • Page ID
    531
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    Recall that

    \[ \int \dfrac{1}{x} dx = \ln |x| + C.\]

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

    \[ \int \dfrac{dx}{1-3x}.\]

    Solution

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

    The integral becomes

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

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

    Larry Green (Lake Tahoe Community College)

    • Integrated by Justin Marshall.