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Math Help - Intregration with Table

  1. #1
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    Intregration with Table

    How do you intergrate dx/x^2(4x^2+9)?

    thanks
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  2. #2
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    Ignore the constants,

    Do you know how to integrate,
    \int \frac{1}{x^2\sqrt{1+x^2}}dx

    Just use the substitution t=\tan^{-1} x.
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  3. #3
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    \int\frac{1}{x^{2}(4x^{2}+9)}dx

    Expand into partial fractions:

    \frac{1}{9}\int\frac{1}{x^{2}}dx-\frac{4}{9}\int\frac{1}{4x^{2}+9}dx

    The first part is easy. You can use a substitution for the second one.

    Try using u=tan^{-1}(\frac{2x}{3}), \;\ du=\frac{6}{4x^{2}+9}dx
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  4. #4
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    \displaystyle \int\frac{dx}{x^2+a^2}=\frac{1}{a}\arctan\frac{x}{  a}+C

    \displaystyle \int\frac{dx}{4x^2+9}=\frac{1}{4}\int\frac{dx}{x^2  +\left(\frac{3}{2}\right)^2}=\frac{1}{4}\cdot\frac  {2}{3}\arctan\frac{2x}{3}+C
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  5. #5
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    Hello, tttcomrader!

    How about Trig Substitution?


    \int\frac{dx}{x^2(4x^2+9)}
    Let 2x \:=\:3\tan\theta\quad\Rightarrow\quad x \:=\:\frac{3}{2}\tan\theta\quad\Rightarrow\quad dx \:=\;\frac{3}{2}\sec^2\!\theta\,d\theta

    . . 4x^2 + 9 \; = \; 9\tan^2\!\theta + 9 \; = \; 9(\tan^2\!\theta + 1) \; = \;9\sec^2\!\theta

    Substitute: . \int\frac{\frac{3}{2}\sec^2\!\theta\,d\theta}{\fra  c{9}{4}\tan^2\!\theta\cdot9\sec^2\!\theta} \;=\;\frac{2}{27}\int\frac{d\theta}{\tan^2\theta} \;= \; \frac{2}{27}\int\cot^2\theta\,d\theta

    . . = \;\frac{2}{27}\int\left(\csc^2\theta - 1\right)\,d\theta \;=\;\frac{2}{27}\left(-\cot\theta - \theta\right) + C\;=\;-\frac{2}{27}(\cot\theta + \theta) + C

    Back-substitute: . -\frac{2}{27}\left[\frac{3}{2x} \,+ \,\arctan\!\left(\frac{2x}{3}\right)\right] + C

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