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Math Help - [SOLVED] Integral and Error Bounds

  1. #1
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    [SOLVED] Integral and Error Bounds

    I can't quite found out how to integrate this
    (dx)/(x + x^(1/3))
    I've tried doing it by partial fraction decomposition and u-sub x^(2/3)
    Can anyone guide me to the way to solve this?

    Also, how do you find K when finding out error bounds for Midpoint/Trapezoid
    I know your suppose to find f''(x) but I don't get what you do after that
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  2. #2
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    nvm found the answers
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  3. #3
    Super Member redsoxfan325's Avatar
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    \int\frac{dx}{x+x^{\frac{1}{3}}}

    Let x = u^3. Thus dx=3u^2\,du.

    Subbing in gives: \int\frac{3u^2}{u^3+u}\,du.

    Factor out the 3 and cancel out a u, which leaves: 3\int\frac{u}{u^2+1}\,du.

    Let t=u^2+1. Thus dt=2u\,du.

    Thus the integral becomes: \frac{3}{2}\int\frac{dt}{t}\,dt.

    This equals \frac{3}{2}ln(t) = \frac{3}{2}ln(u^2+1).

    Subbing back in x^{\frac{1}{3}} for u gives: \frac{3}{2}ln((x^{\frac{1}{3}})^2+1).

    Thus, \int\frac{dx}{x+x^{\frac{1}{3}}} = \frac{3}{2}ln(x^{\frac{2}{3}}+1).


    EDIT: Oh well. Did your answer agree with mine?
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  4. #4
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    yeah I got the same answer but I did it another way

    Factores a x^(1/3) from the bottom and u subbed u = x^(2/3) + 1
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