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Math Help - Need help with definite integral evaluation

  1. #1
    Newbie
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    Need help with definite integral evaluation

    Can someone help me integrate:

    ∫ x^3/X^2+1 dx from 1 to 2

    Thank you
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  2. #2
    Junior Member xterminal01's Avatar
    Joined
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    Once you integrate you get x^2/2+x
    substitude x'es with 2 and the result minus the substitution of 1
    which equals to 5/2 asuming that the "+1" is not in the denominator



    Quote Originally Posted by macadamianchica View Post
    Can someone help me integrate:

    ∫ x^3/X^2+1 dx from 1 to 2

    Thank you
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Apr 2008
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    1,092
    Observe that (x^2 + 1)(x + 1) = x^3 + x^2 + x + 1.

    Therefore,

    \int \frac{x^3}{x^2 + 1}dx =

    \int \frac{x^3 + x^2 + x + 1 - (x^2 + 1) - x}{x^2 + 1}dx =

    \int \frac {(x^2 + 1)(x + 1)}{x^2 + 1} - \frac{x^2 + 1}{x^2 + 1} - \frac {x}{x^2 + 1}dx =

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

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

    \int x dx - \int \frac{x}{x^2 + 1} dx
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  4. #4
    Senior Member
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    Jan 2010
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    Here's an alternative method (you'll get the same result).

    Let u=x^2+1 so that du=2x \, dx.

    Then we have: u-1 = x^2 and \frac{du}{2} = x \, dx.

    So:

    \int_1^2 \frac{x^3}{x^2+1} \, dx ~=~ \int_1^2 \frac{x^2}{x^2+1} \, x \, dx ~=~ \int_2^5 \frac{u-1}{u} \, \frac{du}{2} ~=~ \frac{1}{2} \int_2^5 \left( 1 - \frac{1}{u} \right) du
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