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Thread: integration by substitution

  1. #1
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    integration by substitution

    int(cuberoot(x^3+1)x^5dx
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by samdmansam View Post
    int(cuberoot(x^3+1)x^5dx
    please clarify, you are missing some parenthesis. do you mean $\displaystyle \int \sqrt[3]{x^3 + 1} \cdot x^5~dx$ ?
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  3. #3
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    Int by substitution

    Quote Originally Posted by Jhevon View Post
    please clarify, you are missing some parenthesis. do you mean $\displaystyle \int \sqrt[3]{x^3 + 1} \cdot x^5~dx$ ?
    yes thats exactly what I meant, Thanks
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by samdmansam View Post
    yes thats exactly what I meant, Thanks
    ok, well integration by substitution will work nicely. make a substitution of $\displaystyle u = x^3 + 1$ (and note that $\displaystyle x^5 = x^3 \cdot x^2$)
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  5. #5
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    Make $\displaystyle z^3=x^3+1\implies z^2\,dz=x^2\,dx,$ it's a more direct way to solve the problem, you get rid of the cube root.
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  6. #6
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    Question could you please take it a little further

    my text has very basic examples then skips to questions with cuberoots
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