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Thread: Integrate (2+3x^2)/(x^2(1+x^2))

  1. #1
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    Integrate (2+3x^2)/(x^2(1+x^2))

    $\displaystyle \int \frac {2+3x^2}{x^2(1+x^2)}dx$

    I tried to substitute $\displaystyle 1+x^2$ as 't'. But that failed.
    Then i divided all terms by $\displaystyle x^2$, and that didnt give me any result too..

    Finally, when i saw the answer at the back, i realized that i could write this as a sum of 2 terms..

    But is there any easy method to do this? Without going through all this..?
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  2. #2
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    Re: Integrate (2+3x^2)/(x^2(1+x^2))

    Hello, animesh271094!

    I found two ways . . . Neither is "easy".


    $\displaystyle \int \frac {2+3x^2}{x^2(1+x^2)}dx$

    Partial Fractions

    $\displaystyle \frac{3x^2+2}{x^2(x^2+1)} \;=\;\frac{A}{x} + \frac{B}{x^2} + \frac{Cx}{x^2+1} + \frac{D}{x^2+1}$


    We find that: .$\displaystyle A = 0,\;B = 2,\;C = 0,\;D = 1$


    We have: .$\displaystyle \int\left(\frac{2}{x^2} + \frac{1}{x^2+1}\right)\,dx \;=\; -\frac{2}{x} + \tan^{\text{-}1}\!x + C$


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Substitution

    Let $\displaystyle x = \tan\theta \quad\Rightarrow\quad dx = \sec^2\theta\,d\theta$


    We have: .$\displaystyle \int\frac{2 + 3\tan^2\!\theta}{\tan^2\!\theta\,\sec^2\!\theta}\, \sec^2\!\theta\,d\theta \;=\;\int\frac{2+3\tan^2\!\theta}{\tan^2\!\theta}d \theta$

    . . $\displaystyle =\;\int\left(2\cot^2\!\theta + 3\right)\,d\theta \;=\; \int\left[2(\csc^2\!\theta - 1) + 3\right]\,d\theta $

    . . $\displaystyle =\;\int\left(2\csc^2\!\theta + 1\right)\,d\theta \;=\; -2\cot\theta + \theta + C $


    Back-substitute: .$\displaystyle -2\left(\frac{1}{x}\right) + \tan^{\text{-}1}\!x + C \;=\;-\frac{2}{x} + \tan^{\text{-}1}\!x + C $

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    Re: Integrate (2+3x^2)/(x^2(1+x^2))

    hi animesh271094

    did you try using partial fractions?

    $\displaystyle \frac{2+3x^2}{x^2(1+x^2)}=\frac{A}{x}+\frac{B}{x^2 }+\frac{Cx+D}{1+x^2}$

    just find A,B,c and D.then it will be easier to integrate.
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    Re: Integrate (2+3x^2)/(x^2(1+x^2))

    Thanx a lot guys..
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    Re: Integrate (2+3x^2)/(x^2(1+x^2))

    There is an easier decomposition.
    $\displaystyle \frac{2+3x^2}{x^2(1+x^2)}=\frac{2}{x^2}+\frac{1}{1 +x^2}$
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  6. #6
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    Re: Integrate (2+3x^2)/(x^2(1+x^2))

    hi Plato

    that is the same as Soroban's decomposition.
    Last edited by anonimnystefy; Aug 22nd 2011 at 10:37 AM.
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  7. #7
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    Re: Integrate (2+3x^2)/(x^2(1+x^2))

    Quote Originally Posted by animesh271094 View Post
    $\displaystyle \int \frac {2+3x^2}{x^2(1+x^2)}dx$
    I always avoid partial fractions anytime I can.

    We have $\displaystyle \frac{2+3{{x}^{2}}}{{{x}^{2}}\left( 1+{{x}^{2}} \right)}=\frac{2+2{{x}^{2}}+{{x}^{2}}}{{{x}^{2}}( 1+{{x}^{2}})}=\frac{2\left( 1+{{x}^{2}} \right)+{{x}^{2}}}{{{x}^{2}}\left( 1+{{x}^{2}} \right)}=\frac{2}{{{x}^{2}}}+\frac{1}{1+{{x}^{2}}} .$
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