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Math Help - Tricky integral question..need a little info..

  1. #1
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    Tricky integral question..need a little info..

    Solve: (int) x^2 -x +12/(x^3 + 3x)

    This is what I did so far...

    x3 + 3x = x(x^2 +3)

    = A/x + Bx+C/(x^2 +3)
    = A(x^2+3) + Bx+C(x) = (A+B)x^2 + C(x) + 3A
    A+B)x^2=1
    C=-1
    3A=12

    so A=4, C=-1 and B = -3

    Then....

    4/x = 4(int)1/x -3(int) x/x^2 + 3 -1(int) 1/(x^2 + 3) = it's this part I get confused at...I already know what to do for 4(int)1/x but getting the 1/x^2+3 to be 1/(x^2+1) (so it can be arctan) is my problem as I don't know how to make it into a arctan equation..The answer is 4ln(1/x) - 3/2ln(x^2+3) - arctan (1/sqrt 3/(sqrt 3))..do not understand how they got the 3/2ln(x^2+3) as it looks like you have use the tangent function!
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  2. #2
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    Quote Originally Posted by Jgirl689 View Post
    Solve: (int) x^2 -x +12/(x^3 + 3x)

    This is what I did so far...

    x3 + 3x = x(x^2 +3)

    = A/x + Bx+C/(x^2 +3)
    = A(x^2+3) + Bx+C(x) = (A+B)x^2 + C(x) + 3A
    A+B)x^2=1
    C=-1
    3A=12

    so A=4, C=-1 and B = -3

    Then....

    4/x = 4(int)1/x -3(int) x/x^2 + 3 -1(int) 1/(x^2 + 3) = it's this part I get confused at...I already know what to do for 4(int)1/x but getting the 1/x^2+3 to be 1/(x^2+1) (so it can be arctan) is my problem as I don't know how to make it into a arctan equation..The answer is 4ln(1/x) - 3/2ln(x^2+3) - arctan (1/sqrt 3/(sqrt 3))..do not understand how they got the 3/2ln(x^2+3) as it looks like you have use the tangent function!
    \frac{x^2-x+12}{x(x^2+3)} = \frac{A}{x} + \frac{Bx+C}{x^2+3}<br />

    x^2-x+12 = A(x^2+3) + x(Bx+C)

    matching coefficients ...

    A+B = 1

    C = -1

    3A = 12 ... A = 4 , B = -3



    \int \frac{4}{x} - \frac{3x+1}{x^2+3} \, dx

    \int \frac{4}{x} - \frac{3x}{x^2+3}  - \frac{1}{x^2+3} \, dx

    note that the antiderivative of \frac{1}{x^2+a^2} is \frac{1}{a}\arctan\left(\frac{x}{a}\right) ...

    4\ln|x| - \frac{3}{2}\ln(x^2+3) - \frac{1}{\sqrt{3}}\arctan\left(\frac{x}{\sqrt{3}}\  right) + C
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