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When to use 'ln' in integration?

  1. May 20, 2010 #1
    I get confused when it is 'ok' to use the natural logarithm when integrating a function. As soon as I see a denominator, I am always tempted to simply go 'ln(denominator)/d denominator)' but this is clearly wrong....

    Is it wrong in situations where you have a polynomial denominator? For example

    Integral 1/(x^2 + 2x + 5) dx... would this be ln(x^2 + 2x + 5) / 2x + 2?
     
  2. jcsd
  3. May 20, 2010 #2

    CompuChip

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    You can always check it by differentiating your anti-derivative and checking if you get the integrand back.
    When you have something like
    [tex]\int \frac{1}{2x + 3} \, dx[/tex]
    and you "guess"
    [tex]\frac{\ln |2x + 3|}{2}[/tex]
    you can differentiate and see that it nicely works out (you need the chain rule, which gives a factor of 2 cancelling the denominator).

    However, if you try that for
    [tex]\frac{\ln(x^2 + 2x + 5)}{2x + 2}[/tex]
    you have to use a more complicated rule (e.g. quotient or product + chain rule) to differentiate, you don't just get
    [tex]\frac{1}{x^2 + 2x + 5} \frac{2x + 2}{2x + 2}[/tex]
    but it is followed by "+ .... something you don't want ... "

    So in this case you need to come up with something better. For example, in this quadratic function, you can try "completing the square": if you write
    [tex]\frac{1}{x^2 + 2x + 5} = \frac{1}{(x + a)^2 + b}[/tex]
    you can substitute [itex]u = (x + a) / \sqrt{b}[/itex] and use that
    [tex]\int \frac{1}{1 + u^2} \, du = \operatorname{arctan}(u)[/tex]
     
  4. May 25, 2010 #3
    Thanks for the help!
     
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