Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook