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Thread: Another Substitution indefinite integral problem-just needs checked

  1. #1
    Junior Member
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    Another Substitution indefinite integral problem-just needs checked

    Thanks for looking over it and marking any mistakes!
    \int (\frac{x^2+2}{x^3+6x+3})dx

    u=x^3+6x+3

    du=3x^2+6(dx)

    du=3(x^2+2)dx

    =(x^2+2)dx=\frac{du}{3}

    \int (\frac{1}{3})(u^-1)

    \frac{1}{3}\int (u^-1)

    (\frac{1}{3})(ln{u})+c

    (\frac{1}{3})(ln{x^3+6x+3})+c
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  2. #2
    Super Member
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    Quote Originally Posted by Jim Marnell View Post
    Thanks for looking over it and marking any mistakes!
    \int (\frac{x^2+2}{x^3+6x+3})dx

    u=x^3+6x+3

    du=3x^2+6(dx)

    du=3(x^2+2)dx

    =(x^2+2)dx=\frac{du}{3}

    \int (\frac{1}{3})(u^-1)

    \frac{1}{3}\int (u^-1)

    (\frac{1}{3})(ln{u})+c

    (\frac{1}{3})(ln{x^3+6x+3})+c
    Good work.
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