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Math Help - ∫▒(2x+1)/((x+2)^2) dx A problem with Partial fractions and integration[Attempt to sv]

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    ∫▒(2x+1)/((x+2)^2) dx A problem with Partial fractions and integration[Attempt to sv]

    Hello guys, here is the problem below:

    ∫▒(2x+1)/((x+2)^2) dx A problem with Partial fractions and integration[Attempt to sv]-o5.19.jpg
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    Because you have a case of multiple root in the denominator,
    the correct partial fraction decomposition set-up should be:

    \displaystyle \frac{2x+1}{(x+2)^2} = \frac{A}{x+2}+\frac{B}{(x+2)^2}.
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    MHF Contributor FernandoRevilla's Avatar
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    \dfrac{2x+1}{(x+2)^2}=\dfrac{A}{x+2}+\dfrac{B}{(x+  2)^2}=\dfrac{A(x+2)+B}{(x+2)^2}

    A(x+2)+B=2x+1\Rightarrow A=2,B=-3


    Fernando Revilla


    Edited: Sorry, I didn't see TheCoffeeMachine's post.
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  4. #4
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    I didn't get how you got to this through getting the same denominator: \dfrac{2x+1}{(x+2)^2}=\dfrac{A}{x+2}+\dfrac{B}{(x+ 2)^2}=\dfrac{A(x+2+B}{(x+2)^2} ( The last part)

    Because i get A (x+2)^2 + B(x+2) / (x+2)(x+2)^2
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  5. #5
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    An alternative:

    \displaystyle \int{\frac{2x + 1}{(x + 2)^2}\,dx} = \int{\frac{2x + 4 - 3}{(x + 2)^2}\,dx}

    \displaystyle = \int{\frac{2x + 4}{x^2 + 4x + 4}\,dx} - 3\int{(x + 2)^{-2}\,dx}.


    Substitute \displaystyle u = x^2 + 4x + 4 so that \displaystyle du = (2x + 4)\,dx and substitute \displaystyle v = x + 2 so that \displaystyle dv = dx and the integrals become

    \displaystyle \int{\frac{1}{u}\,du} - 3\int{v^{-2}\,dv}.

    I'm sure you can go from here.
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    thanks
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  7. #7
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Riazy View Post
    I didn't get how you got to this through getting the same denominator: \dfrac{2x+1}{(x+2)^2}=\dfrac{A}{x+2}+\dfrac{B}{(x+ 2)^2}=\dfrac{A(x+2+B}{(x+2)^2} ( The last part)

    Because i get A (x+2)^2 + B(x+2) / (x+2)(x+2)^2

    Divide numerator and denominator by x+2 .


    Fernando Revilla
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