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Math Help - Problem with indefinite integral

  1. #1
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    Problem with indefinite integral

    Hi guys!

    I've got a problem with some indefinite integrals...
    I don't want to solve them using neither the substitution method, nor the parts one, and that's why I'm experiencing some problems...

    So for example, the integral of x/(1+4x^2) = 1/8 ln (4x^2+1), I get it, because (4x^2)=8x... however, I don't get why the integral of 1/(4x^2+1) is 1/2 arctan (2x), why is it 1/2 and why is it 2x?
    The same doubt arises when it is let's say 1/(1+3x^2)...

    I tried using the wolframalpha calculator, but it doesn't show the steps taken in order to achieve the right function...

    I guess it has something to do with 1 being the numerator, instead of x, but even so I don't understand why...

    Thank you guys!
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  2. #2
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    Re: Problem with indefinite integral

    There's an integral formula:

    \frac{1}{a^2 + x^2} = \frac{1}{a}arctanx + C

    Where a is any arbitrary constant.
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: Problem with indefinite integral

    \int \frac{1}{1+4x^2}dx
    =\int \frac{1}{1+(2x)^2}dx
    Let 2x=t \Rightarrow 2dx=dt \Leftrightarrow dx=\fac{1}{2}dt
    thus the integral becomes
    \frac{1}{2} \frac{1}{1+t^2}dt = \frac{1}{2}\arctan(t)+C=\frac{1}{2}\arctan(2x)+C

    Quote Originally Posted by Pupil View Post
    There's an integral formula:
    \frac{1}{a^2 + x^2} = \frac{1}{a}arctanx + C
    Where a is any arbitrary constant.
    This not correct. It should be:
    \int \frac{dx}{a^2+x^2}=\int \frac{dx}{a^2\left(1+\frac{x^2}{a^2}\right)}=\int \frac{dx}{a^2\left[1+\left(\frac{x}{a}\right)^2\right]}
    Let \frac{x}{a}=t \Rightarrow dx=adt thus the integral becomes
     \frac{a}{a^2} \int \frac{dt}{1+t^2}=\frac{1}{a}\arctan(t)+C=\frac{1}{  a}\arctan\left(\frac{x}{a}\right)+C
    Last edited by Siron; May 26th 2012 at 03:46 AM.
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  4. #4
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    Re: Problem with indefinite integral

    Thank you guys!
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