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Math Help - Stuck on an integral

  1. #1
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    Stuck on an integral

    I'm have a bit of trouble with this integral
    \int\frac{dx}{(x+1)(x^2 +4)}
    I'm pretty sure I have to use parts on it but it's been awhile, any help would be appreciated.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by doughnuts View Post
    I'm have a bit of trouble with this integral
    \int\frac{dx}{(x+1)(x^2 +4)}
    I'm pretty sure I have to use parts on it but it's been awhile, any help would be appreciated.
    PFD (Partial Fractions Decomposition)

    \frac{1}{(x+1)(x^2+4)}=\frac{A}{x+1}+\frac{Bx+C}{x  ^2+4}

    Multiplying through by the LCD we get

    1=A(x^2+4)+(Bx+C)(x+1)

    Now if we let x=-1

    we get

    1=A(1+5)+(B(-1)+C)(-1+1)\Rightarrow{A=\frac{1}{5}}

    So now since the other expression has no real roots we must expand

    1=Ax^2+4A+Bx^2+Bx+Cx+C

    Equating coefficients we get

    A+B=0\Rightarrow{B=\frac{-1}{5}}

    and

    4A+C=1\Rightarrow{C=\frac{1}{5}}

    This gives us

    \frac{1}{(x+1)(x^2+4)}=\frac{1}{5}\bigg[\frac{1}{x+1}-\frac{x}{x^2+4}+\frac{1}{x^2+4}\bigg]

    Which implies that

    \int\frac{dx}{(x+1)(x^2+4)}=\frac{1}{5}\int\bigg[\frac{1}{x+1}-\frac{x}{x^2+4}+\frac{1}{x^2+4}\bigg] =\frac{1}{5}\bigg[\ln|x+1|-\frac{1}{2}\ln(x^2+4)+\frac{1}{2}\arctan\left(\fra  c{x}{2}\right)\bigg]+C

    --------------------------------------------------------------------------
    Just out of curiosity how were you planning to use parts

    u=\frac{1}{x+1}
    dv=\frac{1}{x^2+4}?
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    Just out of curiosity how were you planning to use parts

    u=\frac{1}{x+1}
    dv=\frac{1}{x^2+4}?
    Yeah that's what I was trying to use, though now that I see it it looks like partial fractions would be the best way to go, thanks for the speedy response.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by doughnuts View Post
    Yeah that's what I was trying to use, though now that I see it it looks like partial fractions would be the best way to go, thanks for the speedy response.
    Anytime

    Just remember generally if you are given

    \int\frac{N(x)}{D(x)}dx

    Where N(x) and D(x) are polynomials, and D(x) is reducible try PFD
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    Anytime

    Just remember generally if you are given

    \int\frac{N(x)}{D(x)}dx

    Where N(x) and D(x) are polynomials, and D(x) is reducible try PFD
    Check first that deg(N)<deg(D), where "deg" = degree
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  6. #6
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    I hate to bring this thread back from the dead, but after I had this question answered I went on to try and find out if the integral was convergent or divergent from -1 to 6. My answer seemed to be way smaller then it should of been, perhaps I'm doing this wrong.
    I know that in the original equation that X = -1 was a vertical asymptote so I'd should set
    \lim_{t \to -1} \int_t^{6}\frac{dx}{(x+1)(x^2 +4)}
    then
    f(x) = \frac{1}{5}\bigg[\ln|7|-\frac{1}{2}\ln(40)+\frac{1}{2}\arctan\left(3\right  )\bigg]
    Subtract
    \frac{1}{5}\bigg[\ln|0|-\frac{1}{2}\ln(5)+\frac{1}{2}\arctan\left(\frac{-1}{2}\right)\bigg] should equal what the area under the curve is.
    any help would be appreciated.
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  7. #7
    Newbie Msquared's Avatar
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    Quote Originally Posted by doughnuts View Post
    I hate to bring this thread back from the dead, but after I had this question answered I went on to try and find out if the integral was convergent or divergent from -1 to 6. My answer seemed to be way smaller then it should of been, perhaps I'm doing this wrong.
    I know that in the original equation that X = -1 was a vertical asymptote so I'd should set
    \lim_{t \to -1} \int_t^{6}\frac{dx}{(x+1)(x^2 +4)}
    then
    f(x) = \frac{1}{5}\bigg[\ln|7|-\frac{1}{2}\ln(40)+\frac{1}{2}\arctan\left(3\right  )\bigg]
    Subtract
    \frac{1}{5}\bigg[\ln|0|-\frac{1}{2}\ln(5)+\frac{1}{2}\arctan\left(\frac{-1}{2}\right)\bigg] should equal what the area under the curve is.
    any help would be appreciated.

    Ln(0) doesn't exist, the limit as x-> 0 of ln(x) is negative infinity. Hence in the latter portion you would be subtracting this value which would cancel out the negative and make it positive infinity. Hence the improper integral diverges and the area is infinite.

    Other than not pointing that out, you're right.
    Last edited by Msquared; June 21st 2008 at 06:40 PM. Reason: Typo
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