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Math Help - arctan integral

  1. #1
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    arctan integral

    Hi! I just read http://www.mathhelpforum.com/math-he...808-post2.html and the bottom line says
    Quote Originally Posted by jmcq View Post
    The long and short of it is when you have an integral of the form \int\frac{dx}{{a^2+x^2}} you get an result of the form: \frac{1}{a} \arctan \left(\frac{x}{a}\right)+C

    It's a good one to memorize.
    I am trying to do a similar thing but I'm ending up getting the same (wrong) answer all the time.

    \int \frac{dx}{(x+2)^2+4}\\=
    \int \frac{dx}{(\frac{1}{4} \cdot4)\cdot  (x+2)^2+4}\\=
    \frac{1}{4}\int  \frac{dx}{(\frac{1}{2})^2(x+2)^2+1}\\=
    \frac{1}{4}\int  \frac{dx}{(\frac{x+2}{2})^{2}+1}=\frac{1}{4}\left[  arctan(\frac{x+2}{2})+C \right]

    I don't know where the error is, can anyone please show me? It would really appreciated! Please just ask if I've expressed myself unclear!
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  2. #2
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    \int \frac{1}{(x+2)^2 + 4} \, dx

    \int \frac{1}{4\left[\left(\frac{x+2}{2}\right)^2 + 1\right]} \, dx

    \frac{2}{4} \int \frac{\frac{1}{2}}{\left(\frac{x+2}{2}\right)^2 + 1} \,  dx

    \frac{1}{2} \arctan\left(\frac{x+2}{2}\right) + C
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  3. #3
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    Quote Originally Posted by skeeter View Post
    \int \frac{1}{(x+2)^2 + 4} \, dx

    \int \frac{1}{4\left[\left(\frac{x+2}{2}\right)^2 + 1\right]} \, dx

    \frac{2}{4} \int \frac{\frac{1}{2}}{\left(\frac{x+2}{2}\right)^2 + 1} \,  dx

    \frac{1}{2} \arctan\left(\frac{x+2}{2}\right) + C
    Oh : D Of course! Thank you very much!
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