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Math Help - [SOLVED] evaulating the integral- partial fractions

  1. #1
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    [SOLVED] evaulating the integral- partial fractions

    \int_1^e {\frac{{{x^5} + 1}}<br />
{{{x^4} + 2{x^3}}}dx} <br />
    then I divided the two and got:
    \int {x - 2} dx + \int {\frac{{4{x^3}}}<br />
{{{x^4} + 2{x^3}}}dx} <br />
    \frac{{4{x^3}}}<br />
{{({x^3})(x + 2)}} = \frac{{A{x^2} + Bx + C}}<br />
{{{x^3}}} + \frac{D}<br />
{{x + 2}}<br />
    but i am stuck after that
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  2. #2
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    Quote Originally Posted by genlovesmusic09 View Post
    \int_1^e {\frac{{{x^5} + 1}}<br />
{{{x^4} + 2{x^3}}}dx} <br />
    then I divided the two and got:
    \int {x - 2} dx + \int {\frac{{4{x^3}}}<br />
{{{x^4} + 2{x^3}}}dx} <br />
    \frac{{4{x^3}}}<br />
{{({x^3})(x + 2)}} = \frac{{A{x^2} + Bx + C}}<br />
{{{x^3}}} + \frac{D}<br />
{{x + 2}}<br />
    but i am stuck after that
    correction ...

    \frac{x^5+1}{x^4+2x^3} = x - 2 + \frac{4x^3+1}{x^4+2x^3}

    and ...

    \frac{4x^3+1}{x^3(x+2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x+2}<br />
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  3. #3
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    \frac{x^{5}+1}{x^{4}+2x^{3}}=\frac{x^{5}}{x^{3}(x+  2)}+\frac{1}{x^{3}(x+2)}=\frac{x^{2}}{x+2}+\frac{1  }{x^{3}(x+2)}.

    so in order to integrate the first term we have \frac{x^{2}}{x+2}=\frac{x^{2}-4+4}{x+2}=x-2+\frac{4}{x+2}.

    and as for solving \int{\frac{dx}{x^{3}(x+2)}}, put x=\frac1t and the integral becomes -\int{\frac{t^{2}}{1+2t}\,dt}=-\frac{1}{4}\int{\frac{1-\left( 1-4t^{2} \right)}{1+2t}\,dt}.

    now, can you finish this?

    (i definitely dislike partial fractions.)
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  4. #4
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    Quote Originally Posted by Krizalid View Post
    \frac{x^{5}+1}{x^{4}+2x^{3}}=\frac{x^{5}}{x^{3}(x+  2)}+\frac{1}{x^{3}(x+2)}=\frac{x^{2}}{x+2}+\frac{1  }{x^{3}(x+2)}.

    so in order to integrate the first term we have \frac{x^{2}}{x+2}=\frac{x^{2}-4+4}{x+2}=x-2+\frac{4}{x+2}.

    and as for solving \int{\frac{dx}{x^{3}(x+2)}}, put x=\frac1t and the integral becomes -\int{\frac{t^{2}}{1+2t}\,dt}=-\frac{1}{4}\int{\frac{1-\left( 1-4t^{2} \right)}{1+2t}\,dt}.

    now, can you finish this?

    (i definitely dislike partial fractions.)
    Yes - we know!
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  5. #5
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    Quote Originally Posted by skeeter View Post
    correction ...

    \frac{x^5+1}{x^4+2x^3} = x - 2 + \frac{4x^3+1}{x^4+2x^3}

    and ...

    \frac{4x^3+1}{x^3(x+2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x+2}<br />
    then do I use a matrix to solve?
    and if so do you know how to plug it into a ti-83?
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