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Math Help - Help with 2 integrals

  1. #1
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    Help with 2 integrals

    How do I do these 2?

    a) x/(x^2+x-6)

    Tried quotient rule it gave me -x^2 - 6/(x^2+x+6)^2, I don't think it's right

    b) 4/(x^2+4x+4)

    Tried substitution (because theres another similar question between these two on the paper and that was the way to go for it) Don't think my answers worth mentioning
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  2. #2
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    Quote Originally Posted by zombieguy View Post
    How do I do these 2?

    a) x/(x^2+x-6)

    Tried quotient rule it gave me -x^2 - 6/(x^2+x+6)^2, I don't think it's right

    b) 4/(x^2+4x+4)

    Tried substitution (because theres another similar question between these two on the paper and that was the way to go for it) Don't think my answers worth mentioning
    a) \frac{x}{x^2+x-6} = \frac{x}{(x+3)(x-2)} then try partial fractions


    b) \frac{4}{x^2+4x+4} = \frac{4}{(x+2)^2} try a u substitution.
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  3. #3
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    Ok, Thanks

    I've now got:

    a) 3/5ln(x+3)+2/5ln(x-2)

    b) 2/(x+2)ln(x+2)^2

    Is that right?
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  4. #4
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    Quote Originally Posted by zombieguy View Post
    Ok, Thanks

    I've now got:

    a) 3/5ln(x+3)+2/5ln(x-2)

    b) 2/(x+2)ln(x+2)^2

    Is that right?
    The first is (but don't forget the absolute bars), the second isn't. It should be

    \frac{-4}{x+2} + c.
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  5. #5
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    OK, I was able to get your answer for b) but by using:

    u.v-(integral of)(v.du/dx)

    u=4
    du/dx=0 (That gets rid of the integral part)

    v=-(x+2)^-1
    dv/dx= (x+2)^-2

    so: 4.-(x+2)^-1

    It would be nicee if you could show me how to do it your way (using substitution) aswel
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  6. #6
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    \int{\frac{4}{x^2 + 4x + 4}\,dx} = \int{\frac{4}{(x + 2)^2}\,dx}.


    Let u = x + 2 so that \frac{du}{dx} = 1.


    So the integral becomes

    \int{\frac{4}{u^2}\,\frac{du}{dx}\,dx}

     = \int{4u^{-2}\,du}

     = -4u^{-1} + C

     = -\frac{4}{x + 2} + C.
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  7. #7
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    Thank you both for your help
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