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Int (x-2) / (x^2 +2x +3 )

  1. Jan 18, 2006 #1
    does aanybody knows how to do this: int (x-2) / (x^2 +2x +3 )

    pls help..........
     
  2. jcsd
  3. Jan 18, 2006 #2
    does aanybody knows how to do this: int (x-2) / (x^2 +2x +3 )

    write the numerator in the form 1/2[(2x+2)-6]
    now you get on dividing 1/2 { (2x+2)/(x^2+2x+3) - 6/(x^2 +2x +3)}
    for the first part you cand substitute x^2 +2x +3 as t and it will simplify and for the second part make factors (if possible) and use partial fractions or make a perfect square.

    PS this is supposed to be a calculus problem.
     
  4. Jan 18, 2006 #3
    the answer for second part is arctan (x+1) / sqr root 2....
    may i know how do u get it??
     
  5. Jan 18, 2006 #4
    In the second part, in the denominator, you have x^2 +2x+3.

    Write it as (x+1)^2 + 2.

    use the fact that d/dx (arctan(x/a) = 1/(x^2 + a^2)
     
  6. Jan 19, 2006 #5

    VietDao29

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    Homework Helper

    Are you sure that: d/dx (arctan(x/a) = 1/(x^2 + a^2)??? Shouldn't it read:
    [tex]\frac{1}{a} \ \frac{d}{dx} \arctan \left( \frac{x}{a} \right) = \frac{1}{x ^ 2 + a ^ 2}[/tex]?
    [tex]\int \frac{x - 2}{x ^ 2 + 2x + 3} dx = \frac{1}{2} \ \int \frac{2x - 4}{x ^ 2 + 2x + 3} dx = \frac{1}{2} \ \int \frac{2x + 2 - 6}{x ^ 2 + 2x + 3} dx[/tex]
    [tex]= \frac{1}{2} \left( \int \frac{2x + 2}{x ^ 2 + 2x + 3} dx \ - \ \int \frac{6}{(x + 1) ^ 2 + 2} dx \right) = ...[/tex]
    This is just what others have shown you typed in LaTeX.
    Can you go from here?
    Note that:
    [tex]\int \frac{dx}{x ^ 2 + a ^ 2} = \frac{1}{a} \arctan \left( \frac{x}{a} \right) + C[/tex]
     
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