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Math Help - Need help with basic calculus

  1. #1
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    Need help with basic calculus

    How can I integrate the following given?

    int[x^2(1-x^3)/(x^2-1)]dx ans. -1/4x^4-1/2x^2+x-ln(x+1)+c


    int[cos[1/2x]dx/3/4-2/3 sin[1/2x] ans. -3ln(3/4-2/3sin[1/2x]+c


    int=integral symbol


    Thanks
    Last edited by honestliar; August 1st 2009 at 04:49 AM.
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  2. #2
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    Hi honestliar
    1.
    \int \frac{x^2(1-x^3)}{x^2-1}dx ; the numerator divided by denominator will be :

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

    Solve \int \frac{1-x}{x^2-1}dx using partial fraction


    2.
    hint : \frac{d}{dx}sec (x) = tan(x) sex(x)
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by honestliar View Post
    How can I integrate the following given?

    int[x^2(1-x^3)/(x^2-1)]dx

    int[cos[1/2x]dx/3/4-2/3 sin[1/2x]

    int=integral symbol

    Thanks
    \int {\frac{{{x^2}\left( {1 - {x^3}} \right)}}{{{x^2} - 1}}dx} .

    \frac{{{x^2}\left( {1 - {x^3}} \right)}}{{{x^2} - 1}} =  - \frac{{{x^2}\left( {1 - x} \right)\left( {1 + x + {x^2}} \right)}}{{\left( {1 - x} \right)\left( {1 + x} \right)}} =  - \frac{{{x^2}\left( {1 + x + {x^2}} \right)}}{{1 + x}} =  - \frac{{{x^2}\left( {1 + x} \right) + {x^4}}}{{1 + x}} =

    =  - {x^2} - \frac{{{x^4}}}{{1 + x}} =  - {x^2} + \frac{{1 - {x^4} - 1}}{{1 + x}} =  - {x^2} + \frac{{1 - {x^4}}}{{1 + x}} - \frac{1}<br />
{{1 + x}} =

    - {x^2} + \frac{{\left( {1 - {x^2}} \right)\left( {1 + {x^2}} \right)}}{{1 + x}} - \frac{1}{{1 + x}} =  - {x^2} + \frac{{\left( {1 - x} \right)\left( {1 + x} \right)\left( {1 + {x^2}} \right)}}{{1 + x}} - \frac{1}{{1 + x}} =

    =  - {x^2} + \left( {1 - x} \right)\left( {1 + {x^2}} \right) - \frac{1}<br />
{{1 + x}} = 1 - x - {x^3} - \frac{1}{{1 + x}}.

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