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Math Help - partial fractions

  1. #1
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    partial fractions

    I understand how to do these, but I am stuck on how to solve for these 4 variables i got in these equations:
    A + C = 2
    B + D = 0
    A + C = 0
    B + D = 0

    ..just incase i messed up..my original equation is
    integral of (2x^3)/((x^2 + 1))^2
    So i did 2x^3)/(x^2 + 1)(x^2 + 1) = (Ax + B)/(x^2 + 1) +
    (Cx + D)/(X^2 + 1)

    I then got Ax^3 + Ax + Bx^2 + B + Cx^3 + Cx + Dx^2 + D
    Then I got those 4 equations above...I am not seeing how to solve for the variables right now..maybe a brain fart.
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  2. #2
    Senior Member tukeywilliams's Avatar
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     \frac{2x^3}{(x^2+1)^2} = \frac{Ax+b}{x^2+1} + \frac{Cx+d}{(x^2+1)^2} . In the second term you forgot to square it.
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  3. #3
    Eater of Worlds
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    Doing the PF thing:

    \frac{Ax+B}{x^{2}+1}+\frac{Cx+D}{(x^{2}+1)^{2}}=2x  ^{3}

    Multiply by the LCD (x^{2}+1)^{2}, expand out and we get:

    Ax^{3}+Bx^{2}+Ax+B+Cx+D=2x^{3}

    Equate coefficients:

    A=2
    B=0
    A+C=0
    B+D=0

    Two of the vaiables are already solved, so

    solve this system and we can see A=2, B=0, C=-2, D=0

    So, you have:

    \frac{2x}{x^{2}+1}-\frac{2x}{(x^{2}+1)^{2}}
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  4. #4
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    Quote Originally Posted by tukeywilliams View Post
     \frac{2x^3}{(x^2+1)^2} = \frac{Ax+b}{x^2+1} + \frac{Cx+d}{(x^2+1)^2} . In the second term you forgot to square it.

    I am getting equations that make no sense now..

    Ax^3 + Ax + Bx^2 + B + Cx^5 + 2Cx^3 + Cx + Dx^4 + 2Dx^2 + D
    so..
    for the x^5 terms: C = 0
    x^4 terms: D = 0
    X^3 terms: A + 2C = 2 A = 2 since C = 0
    X^2 terms: B + 2D = 0 B = 0 since D = 0
    X^1 terms: A + C = 0 (impossible since A = 2)
    X^0 terms: B + D = 0
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  5. #5
    Eater of Worlds
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    Quote Originally Posted by davecs77 View Post
    I am getting equations that make no sense now..

    Ax^3 + Ax + Bx^2 + B + Cx^5 + 2Cx^3 + Cx + Dx^4 + 2Dx^2 + D
    so..
    for the x^5 terms: C = 0
    x^4 terms: D = 0
    X^3 terms: A + 2C = 2 A = 2 since C = 0
    X^2 terms: B + 2D = 0 B = 0 since D = 0
    X^1 terms: A + C = 0 (impossible since A = 2)<---what if C=-2
    X^0 terms: B + D = 0
    I think you have an oversight.
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  6. #6
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by davecs77 View Post
    I understand how to do these, but I am stuck on
    integral of (2x^3)/((x^2 + 1))^2
    Just in case, there's no necessity to apply partial fractions.

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

    Set u=x^2+1\implies{du}=2x~dx, therefore

    \begin{aligned}<br />
\int {\frac{{2x^3 }}<br />
{{(x^2 + 1)^2 }}~dx} &= \int {\frac{{u - 1}}<br />
{{u^2 }}~du}\\<br />
&= \ln \left| u \right| + \frac{1}<br />
{u} + k\\<br />
&= {\color{blue}\ln \left(x^2 + 1\right) + \frac{1}<br />
{{x^2 + 1}} + k},~~k\in\mathbb{R}~\blacksquare<br />
\end{aligned}
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