Skip to main content
Mathematics LibreTexts

4.2: Logs and Integrals

  • Page ID
    531
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    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


    This page titled 4.2: Logs and Integrals is shared under a not declared license and was authored, remixed, and/or curated by Larry Green.

    • Was this article helpful?