Skip to main content
Mathematics LibreTexts

7.4: Homework- u-substitution

  • Page ID
    88687
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    1. As a review of the chain rule from derivatives, find \(\frac{d}{dx} e^{-x^2 + 1}\).
      \(2x e^{x^2 + 1}\)
      ans
    2. Read through section 6A again and then read through section 6B.
    3. Watch the Khan Academy video u-substitution.
    4. Compute the following indefinite integral using the method of \(u\)-substitution.

      \(\int (5x^4 + 2x) e^{x^5 + x^2}dx\)

      \(e^{x^5 + x^2} + C\)
      ans
    5. Watch another example of \(u\)-substitution: u-substitution 2.
    6. Compute the following indefinite integral using the method of \(u\)-substitution.

      \(\int \frac{2x}{x^2 - 5}dx\)

      \(\ln(x^2 - 5) + C\)
      ans
    7. Reread the part about the chain rule shortcut for \(u\)-substitution in chapter 6 of the online notes, and reread Example 6B.2. Then try the following problems.
      1. \(\int e^{-3x}dx\).
        \(\frac{1}{-3} e^{-3x} + C\)
        ans
      2. \(\int \left( \frac{1}{2}x - 1 \right)^4dx\)
        \(.4 \left( \frac{1}{2} x - 1 \right)^5 + C\)
        ans

        .

      3. \(\int_0^2 (5x - 3)^3dx\).
        \(116\)
        ans
      4. \(\int \frac{1}{7x - 2}dx\)
        \(\frac{1}{7} \ln(7x - 2) + C\)
        ans
      5. \(\int \frac{1}{\sqrt{0.5x - 1}}dx\)
        \(4 \sqrt{0.5x - 1} + C\)
        ans
    8. Try some more \(u\)-substitution integrals.
      1. \(\int (8x^3) (x^4 + 1)^2dx\)
        1. We see that \(u = x^4 + 1\) in this case.
        2. We see \(\frac{du}{dx} = 4x^3\), so \(dx = \frac{1}{4x^3} du\).
        3. We have

          \[\begin{align*} & = \int (8x^3) u^2 \frac{1}{4x^3} du \\ & = \int 8 u^2 \frac{1}{4}du \\ & = 2 \int u^2du \end{align*}\]

        4. Integrating we have

          \(2 \int u^2du = 2 \left( \frac{1}{3} u^3 \right ) + C = \frac{2}{3} u^3 + C\)

        5. Substitution of \(u = x^4 + 1\) yields

          \(\frac{2}{3} (x^4 + 1)^3 + C\)

        ans
      2. \(\int (x^4 + 1)^2 (8x^3)dx\)
        This is exactly the same as the previous problem, just written a different way. No need to redo work.
        ans
      3. \(\int (3x^2 - 1) e^{x^3 - x}dx\)
        1. We see that \(u = x^3 - x\) in this case.
        2. We see \(\frac{du}{dx} = 3x^2 - 1\), so \(dx = \frac{1}{3x^2 - 1} du\).
        3. We have

          \[\begin{align*} & = \int (3x^2 - 1) e^u \frac{1}{3x^2 - 1} du \\ & = \int e^udu \end{align*}\]

        4. Integrating we have

          \(\int e^udu = e^u + C\)

        5. Substitution of \(u = x^3 - x\) yields

          \(e^{x^3 - x} + C\)

        ans
      4. \(\int (x^2 - \frac{1}{3}) e^{x^3 - x}dx\)
        \(\frac{1}{3} e^{x^3 - x} + C\)
        ans
      5. \(\int (e^x+ x)^5 (e^x+1)dx\)
        \(\frac{1}{6} (e^x + x)^6 + C\)
        ans
      6. \(\int \frac{2x}{x^2 - 1}dx\)
        \(\ln(x^2 -1) + C\)
        ans
      7. \(\int e^x \sqrt{e^x + 1}dx\)
        ans
      8. \(\int \frac{\sin(x)}{\cos(x)}dx\)
        \(-\ln(\cos(x)) + C\)
        ans
      9. \(\int \frac{\sqrt{ \ln(x) }}{x}dx\)
        ans
      10. \(\int \frac{x}{(x^2 - 5)^3}dx\)
        1. We see that \(u = x^2 - 5\) in this case.
        2. We see \(\frac{du}{dx} = 2x\), so \(dx = \frac{1}{2x} du\).
        3. We have

          \[\begin{align*} & = \int \frac{x}{u^3}\frac{1}{2x}du \\ & = \int \frac{1}{2u^3}du \\ & = \int \frac{1}{2}u^{-3}du \end{align*}\]

        4. Integrating we have

          \(\int \frac{1}{2}u^{-3}du = - \frac{1}{4} u^{-2} + C\)

        5. Substitution of \(u = x^2 - 5\) yields

          \(- \frac{1}{4} (x^2 - 5)^{-2} + C\)

        ans
      11. \(\int \frac{\cos(x)}{\sqrt{\sin(x) + 1}}dx\)
        \(2\sqrt{\sin(x) + 1} + C\)
        ans
      12. \(\int \frac{1}{x (2 \ln(x) + 1)^4}dx\).
        \(-\frac{1}{6(2 \ln(x) + 1)^3} + C\)
        ans
      13. \(\int (e^{5x} + 1)^9 (e^{5x} )dx\).
        \(\frac{1}{2} (e^{5x} + 1)^{10} + C\)
        ans

    This page titled 7.4: Homework- u-substitution is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Tyler Seacrest via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

    • Was this article helpful?