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Math Help - Where have I gone wrong?

  1. #1
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    Where have I gone wrong?

    Hi we were asked to solve the following integral by substitution
    \int_{}^{}\frac{x+1}{2x^2+4x}dx
    I first took a factor of 1/2 out, to give
    \frac{1}{2}\int_{}^{}\frac{x+1}{x^2+2x}dx
    I then let
    u=x^2+2x
    so
    \frac{du}{dx}=2x+2

    \frac{1}{2}\frac{du}{dx}=x+1
    substituting for u and du/dx gives
    \frac{1}{2}\int_{}^{}\frac{\frac{1}{2}\frac{du}{dx  }}{u}dx

    \frac{1}{2}\int_{}^{}\frac{1}{2}\frac{1}{u}\frac{d  u}{dx}dx

    \frac{1}{4}\int_{}^{}\frac{1}{u}du

    \frac{1}{4}ln\left|u\right|+c

    \frac{1}{4}ln\left|x^2+2x\right|+c
    Anything wrong there?
    Because if I integrate WITHOUT taking the first 1/2 factor out, leaving u as 2x^2 + 4x, i get the same answer but with 2x^2 + 4x inside the ln. These answers are surely different, yet both make sense so what's gone wrong?
    I've run this past my classmate and teacher and none of use can find the mistake, so any help is greatly appreciated.
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  2. #2
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    Quote Originally Posted by chug1 View Post
    Hi we were asked to solve the following integral by substitution
    \int_{}^{}\frac{x+1}{2x^2+4x}dx
    I first took a factor of 1/2 out, to give
    \frac{1}{2}\int_{}^{}\frac{x+1}{x^2+2x}dx
    I then let
    u=x^2+2x
    so
    \frac{du}{dx}=2x+2

    \frac{1}{2}\frac{du}{dx}=x+1
    substituting for u and du/dx gives
    \frac{1}{2}\int_{}^{}\frac{\frac{1}{2}\frac{du}{dx  }}{u}dx

    \frac{1}{2}\int_{}^{}\frac{1}{2}\frac{1}{u}\frac{d  u}{dx}dx

    \frac{1}{4}\int_{}^{}\frac{1}{u}du

    \frac{1}{4}ln\left|u\right|+c

    \frac{1}{4}ln\left|x^2+2x\right|+c
    Anything wrong there?
    Because if I integrate WITHOUT taking the first 1/2 factor out, leaving u as 2x^2 + 4x, i get the same answer but with 2x^2 + 4x inside the ln. These answers are surely different, yet both make sense so what's gone wrong?
    I've run this past my classmate and teacher and none of use can find the mistake, so any help is greatly appreciated.
    Both answers are correct.

    Note that \frac{1}{4} \ln |2x^2 + 4x| + K = \frac{1}{4} \ln |2(x^2 + 2x)| + K = \frac{1}{4} ( \ln |x^2 + 2x| + \ln 2 ) + K

     = \frac{1}{4}\ln |x^2 + 2x| + \frac{1}{4} \ln (2) + K = \frac{1}{4}\ln |x^2 + 2x| + C

    where C is just as arbitrary as K.
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  3. #3
    Junior Member
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    Ah ha! Thanks heaps!
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