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Integration problem.

  1. Jan 10, 2010 #1
    I understand the mechanics of how this happens but i don't really understand why.

    [tex]\frac{a}{b}\int\frac{1}{x+c}dx\neq\int\frac{a}{b(x+c)}dx[/tex]

    Why can't the constant be taken out?:confused:
     
  2. jcsd
  3. Jan 10, 2010 #2
    What makes you think it can't be?
     
  4. Jan 10, 2010 #3
    This:
    [tex]\frac{a}{b}\int\frac{1}{x+c}dx=\frac{a}{b}ln(x+c)+C[/tex]

    [tex]\int\frac{a}{b(x+c)}dx=\frac{a}{b}ln(bx+bc)+C[/tex]
     
  5. Jan 10, 2010 #4
    Consider
    [tex]
    \int 5x
    [/tex]

    So, which is right?

    [tex]
    \int 5x = \frac{(5x)^2}{2} + C
    [/tex]
    or
    [tex]
    \int 5x = 5\frac{x^2}{2} + C
    [/tex]
    Even in regular integration, you always pull off the constants. Just because you have 1/x doesn't mean the constant shouldn't be pulled out.

    However, it does worth mentioning that both your answers are actually right.

    [tex]
    \frac{a}{b}ln(bx+bc)+C = \frac{a}{b}ln(b(x+c))+C = \frac{a}{b}ln(x+c)+ \frac{a}{b}ln b + C = \frac{a}{b}ln(x+c)+D
    [/tex]
    Where D is just another constant.
     
  6. Jan 10, 2010 #5
    I know.
    Ok now I understand it. I actually run into this problem while trying to calculate integration factors for ODEs. This should simplify my calculations.

    Thank you.
     
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