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Thread: help plz

  1. #1
    Junior Member
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    help plz

    How would you solve this
    integral of dx/(1+4x^2)
    I know arctan is 1+x^2, but dont see how to solve this to make it look like arctan. thanks.
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  2. #2
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    $\displaystyle 4x^2 = (2x)^2$
    So use the substitution $\displaystyle t=2x$.
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    $\displaystyle
    \begin{aligned}
    \int\frac1{1+4x^2}\,dx&=\frac14\int\frac1{\left(\d frac12\right)^2+x^2}\\&=\frac12\arctan2x+k,\,k\in\ mathbb R
    \end{aligned}
    $

    Where $\displaystyle \color{red}\int\frac1{a^2+u^2}\,du=\frac1a\arctan\ frac ua+k,\,a\ne0$
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  4. #4
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    Quote Originally Posted by Krizalid View Post
    $\displaystyle
    \begin{aligned}
    \int\frac1{1+4x^2}\,dx&=\frac14\int\frac1{\left(\d frac12\right)^2+x^2}\\&=\frac12\arctan2x+k,\,k\in\ mathbb R
    \end{aligned}
    $

    Where $\displaystyle \color{red}\int\frac1{a^2+u^2}\,du=\frac1a\arctan\ frac ua+k,\,a\ne0$
    Krizalid, it should be $\displaystyle |a|$.
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