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Thread: Indefinite Integral Polynomial in Denominator: integrate (2x/(x^2 + 6x + 13)) dx

  1. #1
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    Question Indefinite Integral Polynomial in Denominator: integrate (2x/(x^2 + 6x + 13)) dx

    So I'm trying to integrate the indefinite integral of (2x/(x^2 + 6x + 13)) and I get to an answer that I know doesn't work but I'm only off but a very small amount and I'd like to know where I went wrong. I'll post my work and the incorrect answer below:


    indefinite integral of ( (2x/(x^2 + 6x + 13)) dx)
    indefinite integral of ( (2x/((x+3)^2 + 4)) dx)

    Let u = x + 3

    indefinite integral of ( (2(u-3)/(4 + u^2)) du)

    (1/2) * indefinite integral of ( ((u-3)/(1 + (u/2)^2)) du)

    (1/2) * ( indefinite integral of ( (u/(1 + (u/2)^2)) du) - indefinite integral of ( (3/(1 + (u/2)^2)) du) )

    Let z = 1 + (u/2)^2

    indefinite integral of (dz/z) - (3/2) * indefinite integral of ( (1/(1 + (u/2)^2)) du)

    ln( (x^2 + 6x + 13)/4 ) - (3/2) * arctan( (x+3)/2 ) + C


    The correct answer should be:

    ln(x^2 + 6x + 13) - 3*arctan( (x+3)/2 ) + C

  2. #2
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    Quote Originally Posted by debased View Post
    So I'm trying to integrate the indefinite integral of (2x/(x^2 + 6x + 13)) and I get to an answer that I know doesn't work but I'm only off but a very small amount and I'd like to know where I went wrong. I'll post my work and the incorrect answer below:


    indefinite integral of ( (2x/(x^2 + 6x + 13)) dx)
    indefinite integral of ( (2x/((x+3)^2 + 4)) dx)

    Let u = x + 3

    indefinite integral of ( (2(u-3)/(4 + u^2)) du)

    (1/2) * indefinite integral of ( ((u-3)/(1 + (u/2)^2)) du)

    (1/2) * ( indefinite integral of ( (u/(1 + (u/2)^2)) du) - indefinite integral of ( (3/(1 + (u/2)^2)) du) )

    Let z = 1 + (u/2)^2 -->dz = udu/2 or 2dz = udu. You replaced udu with dz which is not true

    indefinite integral of (dz/z) - (3/2) * indefinite integral of ( (1/(1 + (u/2)^2)) du)

    ln( (x^2 + 6x + 13)/4 ) - (3/2) * arctan( (x+3)/2 ) + C


    The correct answer should be:

    ln(x^2 + 6x + 13) - 3*arctan( (x+3)/2 ) + C
    dz = udu/2 or 2dz = udu. You replaced udu with dz which is not true. Also, ln( (x^2 + 6x + 13)/4 ) = ln((x^2 + 6x + 13) - ln(4) and ln(4) can be absorbed by C.
    Fix these mistakes and see what you get.
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

  3. #3
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    Quote Originally Posted by debased View Post
    So I'm trying to integrate the indefinite integral of (2x/(x^2 + 6x + 13)) and I get to an answer that I know doesn't work but I'm only off but a very small amount and I'd like to know where I went wrong. I'll post my work and the incorrect answer below:


    indefinite integral of ( (2x/(x^2 + 6x + 13)) dx)
    indefinite integral of ( (2x/((x+3)^2 + 4)) dx)

    Let u = x + 3

    indefinite integral of ( (2(u-3)/(4 + u^2)) du)

    (1/2) * indefinite integral of ( ((u-3)/(1 + (u/2)^2)) du)

    (1/2) * ( indefinite integral of ( (u/(1 + (u/2)^2)) du) - indefinite integral of ( (3/(1 + (u/2)^2)) du) )

    Let z = 1 + (u/2)^2

    indefinite integral of (dz/z) - (3/2) * indefinite integral of ( (1/(1 + (u/2)^2)) du)

    ln( (x^2 + 6x + 13)/4 ) - (3/2) * arctan( (x+3)/2 ) + C


    The correct answer should be:

    ln(x^2 + 6x + 13) - 3*arctan( (x+3)/2 ) + C

    [tex][f(x)]^{2} + A^{2}[/tex] typically, substitute f(x) = A*tan(u)

    [tex][f(x)]^{2} - A^{2}[/tex] typically, substitute f(x) = A*sin(u)

    [tex][A^{2} - f(x)]^{2}[/tex] typically, substitute f(x) = A*sec(u)

    Let's see what you get.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  4. #4
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    Quote Originally Posted by Jomo View Post
    dz = udu/2 or 2dz = udu. You replaced udu with dz which is not true. Also, ln( (x^2 + 6x + 13)/4 ) = ln((x^2 + 6x + 13) - ln(4) and ln(4) can be absorbed by C.
    Fix these mistakes and see what you get.
    I didn't replace udu with dz I did substitute in 2dz I just skipped the step of showing that the 1/2 constant in front of that integral would cancel with the 2 you pull out from 2dz. I fixed the ln(4) mistake but I'm still left with

    ln((x^2 + 6x + 13) - (3/2) * arctan((x+3)/2) + C

  5. #5
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    Quote Originally Posted by debased View Post
    ... - indefinite integral of ( (3/(1 + (u/2)^2)) du) )

    ... - (3/2) * indefinite integral of ( (1/(1 + (u/2)^2)) du)

    ... - (3/2) * arctan( (x+3)/2 ) + C

    The correct answer should be:

    ... - 3*arctan( (x+3)/2 ) + C
    Quote Originally Posted by debased View Post
    I didn't replace udu with dz I did substitute in 2dz I just skipped the step of showing that the 1/2 constant in front of that integral would cancel with the 2 you pull out from 2dz. I fixed the ln(4) mistake but I'm still left with

    ln((x^2 + 6x + 13) - (3/2) * arctan((x+3)/2) + C
    Focus on the one integral I singled out above. This is where you missed something in the substitution; you didn't show the details of what you did for this one. Presumably you made a substitution like z = u/2, so du = 2 dz, and should get

    INT (1/(1 + (u/2)^2)) du = INT (1/(1 + z^2)) *2 dz = 2 arctan(z) + C

    This leads to the correct answer.

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    Quote Originally Posted by Dr.Peterson View Post
    Focus on the one integral I singled out above. This is where you missed something in the substitution; you didn't show the details of what you did for this one. Presumably you made a substitution like z = u/2, so du = 2 dz, and should get

    INT (1/(1 + (u/2)^2)) du = INT (1/(1 + z^2)) *2 dz = 2 arctan(z) + C

    This leads to the correct answer.
    I thought the answer should be a 3 arctan(z) + C; aren't you forgetting to distribute the 1/2 that I pulled out before separating it into two integrals?

    (1/2) * ( indefinite integral of ( (u/(1 + (u/2)^2)) du) - indefinite integral of ( (3/(1 + (u/2)^2)) du) )

    Let z = 1 + (u/2)^2

    (1/2) * indefinite integral of (2dz/z) - (1/2) * indefinite integral of ( (3/(1 + (u/2)^2)) du)

    indefinite integral of (dz/z) - (3/2) * indefinite integral of ( (1/(1 + (u/2)^2)) du)

  7. #7
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    Quote Originally Posted by debased View Post
    I thought the answer should be a 3 arctan(z) + C; aren't you forgetting to distribute the 1/2 that I pulled out before separating it into two integrals?

    (1/2) * ( indefinite integral of ( (u/(1 + (u/2)^2)) du) - indefinite integral of ( (3/(1 + (u/2)^2)) du) )

    Let z = 1 + (u/2)^2

    (1/2) * indefinite integral of (2dz/z) - (1/2) * indefinite integral of ( (3/(1 + (u/2)^2)) du)

    indefinite integral of (dz/z) - (3/2) * indefinite integral of ( (1/(1 + (u/2)^2)) du)
    I didn't forget the 1/2; I was dealing with the integral itself, leaving the finishing for you. But you seem to have lost the 2:

    I showed that indefinite integral of ( (1/(1 + (u/2)^2)) du) = 2 arctan(z); when you multiply that by 3/2, you're left with just what you want.

  8. #8
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    Quote Originally Posted by Dr.Peterson View Post
    I didn't forget the 1/2; I was dealing with the integral itself, leaving the finishing for you. But you seem to have lost the 2:

    I showed that indefinite integral of ( (1/(1 + (u/2)^2)) du) = 2 arctan(z); when you multiply that by 3/2, you're left with just what you want.
    Thanks, your help in the other thread got me to realize I need to use another substitution since you can't have the (u/2)^2 go directly to an arctan. Got the answer now, much appreciated!

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