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Math Help - integration by partial fractions

  1. #1
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    integration by partial fractions

    I am having some trouble with \int\frac{x+1}{x^2+4x+3}~dx. I can break it down to \frac{x+1}{x^2+4x+3}=\frac{A}{x+1}+\frac{B}{x+3}. Then let x=-3, and get B=1. Then let x=-1, and get A=0. A=0 will not work because it will then integrate to ln lx+3l + C and that will not differentiate to what I had originally. However, \int\frac{x+1}{x^2+4x+3}~dx=\int\frac{x+1}{(x+1)(x  +3)}~dx=\int\frac1{x+3}~dx. And therefore is this what I am really integrating?

    Just making sure my logic is correct.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Possible actuary View Post
    I am having some trouble with \int\frac{x+1}{x^2+4x+3}~dx. I can break it down to \frac{x+1}{x^2+4x+3}=\frac{A}{x+1}+\frac{B}{x+3}. Then let x=-3, and get B=1. Then let x=-1, and get A=0. A=0 will not work because it will then integrate to ln lx+3l + C and that will not differentiate to what I had originally. However, \int\frac{x+1}{x^2+4x+3}~dx=\int\frac{x+1}{(x+1)(x  +3)}~dx=\int\frac1{x+3}~dx. And therefore is this what I am really integrating?

    Just making sure my logic is correct.
    you are correct. the last factorization method is what i would use. don't worry that things don't look exactly the same algebraically. what we actually did was take 1/(x + 3) and multiply it by 1, namely by (x + 1)/(x + 1).

    if you plot a graph of (x + 1)/(x^2 + 4x + 3), you would get exactly the same graph as the graph for 1/(x + 3)
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jhevon View Post
    you are correct. the last factorization method is what i would use. don't worry that things don't look exactly the same algebraically. what we actually did was take 1/(x + 3) and multiply it by 1, namely by (x + 1)/(x + 1).

    if you plot a graph of (x + 1)/(x^2 + 4x + 3), you would get exactly the same graph as the graph for 1/(x + 3)
    But make sure your integration region doesn't include x = -1! Because \frac{x +1}{x^2 + 4x + 3} doesn't exist at x = -1. \frac{1}{x + 3} \neq \frac{x +1}{x^2 + 4x + 3}. The domains of the two expressions are different.

    -Dan
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