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Thread: Integrals

  1. #1
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    Exclamation Integrals

    Hi if you integrate (9x^2-2x)/(6x^3-2x^2), what's the answer? The textbook says 1/2 * log(6x^3-2x^2) + c...is that correct? Because I keep on getting 3/4 * log (6x^2 - 2x) + c

    PLEASE HELP!!!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by xwrathbringerx View Post
    Hi if you integrate (9x^2-2x)/(6x^3-2x^2), what's the answer? The textbook says 1/2 * log(6x^3-2x^2) + c...is that correct? Because I keep on getting 3/4 * log (6x^2 - 2x) + c

    PLEASE HELP!!!
    Note that $\displaystyle \int\frac{9x^2-2x}{6x^3-2x^2}\,dx=\frac{1}{2}\int\frac{9x^2-2x}{3x^3-x^2}\,dx$

    So if $\displaystyle u=3x^3-x^2$, $\displaystyle \,du=9x^2-2x\,dx$

    So, $\displaystyle \frac{1}{2}\int\frac{9x^2-2x}{3x^3-x^2}\,dx\xrightarrow{u=3x^3-x^2}{}\frac{1}{2}\int\frac{\,du}{u}$.

    Even though this result and the back of the book's result are different, they're actually the same! (can you figure out why?)
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  3. #3
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    Quote Originally Posted by xwrathbringerx View Post
    Hi if you integrate (9x^2-2x)/(6x^3-2x^2), what's the answer? The textbook says 1/2 * log(6x^3-2x^2) + c...is that correct? Because I keep on getting 3/4 * log (6x^2 - 2x) + c

    PLEASE HELP!!!
    Note that if you let $\displaystyle u = 6x^3 - 2x^2$, then $\displaystyle \frac{du}{dx} = 18x^2 - 4x$.


    You can rearrange the integrand so that

    $\displaystyle \int{\frac{9x^2 - 2x}{6x^3 - 2x^2}\,dx} = \frac{1}{2}\int{\frac{1}{6x^3 - 2x^2}\cdot (18x^2 - 4x)\,dx}$

    $\displaystyle = \frac{1}{2}\int{\frac{1}{u}\,\frac{du}{dx}\,dx}$

    $\displaystyle = \frac{1}{2}\int{\frac{1}{u}\,du}$

    $\displaystyle = \frac{1}{2}\ln{|u|} + C$

    $\displaystyle = \frac{1}{2}\ln{|6x^3 - 2x^2|} + C$.


    The book is correct.
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