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Math Help - Another Integral

  1. #1
    MHF Contributor Amer's Avatar
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    Another Integral

    \displaystyle \int_{0}^{\infty} \frac{2x^2+3}{x^4+7x^2+1} dx

    any hints
    I tried much I tried by parts considering dx=dv u=f(x) failed
    I tried sub u=x^2 failed
    parts dv=2x^2+3 u=1/(x^4+7x^2+1) failed and another things but I failed at all

    Thanks
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  2. #2
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    \displaystyle\int_0^{\infty}\frac{2x^2}{x^4+7x^2+1  } \ dx

    \displaystyle u=\tan^{-1}\left(\frac{(\sqrt{5}-3)x}{2}\right)

    \displaystyle 3\int_0^{\infty}\frac{1}{x^4+7x^2+1} \ dx

    \displaystyle u_2=\tan^{-1}\left(\frac{(\sqrt{5}+3)x}{2}\right)
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  3. #3
    MHF Contributor Amer's Avatar
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    can I know how u choose this sub specially the constant multiply with x in the arctan
    although it is still messy :S
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  4. #4
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    Quote Originally Posted by Amer View Post
    can I know how u choose this sub specially the constant multiply with x in the arctan
    although it is still messy :S
    I looked at the solutions to the separate indefinite integrals. There is probably an easier way to do it though.
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  5. #5
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    Quote Originally Posted by Amer View Post
    \displaystyle \int_{0}^{\infty} \frac{2x^2+3}{x^4+7x^2+1} dx

    any hints
    I tried much I tried by parts considering dx=dv u=f(x) failed
    I tried sub u=x^2 failed
    parts dv=2x^2+3 u=1/(x^4+7x^2+1) failed and another things but I failed at all

    Thanks

    Unless there's some snaky trick, this seems to be a specially nasty exercise in decomposition of rational functions:

    \displaystye{x^4+7x^2+1=\left(x^2+\frac{7-3\sqrt{5}}{2}\right)\left(x^2+\frac{7+3\sqrt{5}}{2  }\right)} , so :

    \displaystyle{\frac{2x^2+3}{x^4+7x^2+1}=\frac{Ax+B  }{x^2+\frac{7-3\sqrt{5}}{2}}+\frac{Cx+D}{x^2+\frac{7+3\sqrt{5}}{  2}} , and now multiply

    by the common denominator, choose nice values of x in both sides and find the coefficients

    A,B,C,D. It looks horrible, indeed...but who knows? Perhaps there's another way...

    Tonio
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  6. #6
    MHF Contributor Amer's Avatar
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    Thanks I will work on it

    \displaystyle{\frac{2x^2+3}{x^4+7x^2+1}=\frac{Ax+B  }{x^2+\frac{7-3\sqrt{5}}{2}}+\frac{Cx+D}{x^2+\frac{7+3\sqrt{5}}{  2}}

    I found that

    B = \frac{3\sqrt{5} - 4}{3\sqrt{5}}

    D = \frac{3\sqrt{5} +4}{3\sqrt{5}}

    \displaystyle \int_{0}^{\infty} \frac{\frac{3\sqrt{5} - 4}{3\sqrt{5}}}{x^2 + \frac{7-3\sqrt{5}}{2}} + \frac{\frac{3\sqrt{5} +4}{3\sqrt{5}}}{x^2 + \frac{7+3\sqrt{5}}{2}} dx

    \displaystyle \int_{0}^{\infty} \frac{\frac{3\sqrt{5} - 4}{3\sqrt{5}}}{x^2 + \frac{7-3\sqrt{5}}{2}} dx + \int_{0}^{\infty} \frac{\frac{3\sqrt{5} +4}{3\sqrt{5}}}{x^2 + \frac{7+3\sqrt{5}}{2}} dx

    both are arctan thanks
    Last edited by Amer; February 8th 2011 at 12:24 PM.
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  7. #7
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    Write


     \displaystyle  2x^2 + 3 = A(x^2 - 1) + B(x^2 + 1) where  A,B are constants , solving it we obtain  \displaystyle   A = -0.5 ~,~ B = 2.5

    We have

     \displaystyle  \int \frac{2x^2 + 3}{x^4 + 7x^2 + 1}~dx

     \displaystyle  = -\frac{1}{2} \int \frac{x^2 - 1}{x^4 + 7x^2 + 1 }~dx  + \frac{5}{2} \int \frac{x^2 + 1}{x^4 + 7x^2 + 1}~dx

    For the first integral , consider

     \displaystyle   \int \frac{1 - 1/x^2}{x^2 + 7 + 1/x^2 }~dx

     \displaystyle  = \int \frac{d(x+1/x)}{(x+1/x)^2 + 5 }

    Sub. \displaystyle   u = x+1/x

    \displaystyle   = \int \frac{du}{u^2+5} = \frac{1}{\sqrt{5}} \tan^{-1}(u/\sqrt{5})

     \displaystyle  = \frac{1}{\sqrt{5}} \tan^{-1}\left( \frac{x^2+1}{\sqrt{5} x } \right)

    For the second integral , you may find it is a similar trick

    \displaystyle    \int \frac{1 + 1/x^2}{x^2 + 7 + 1/x^2 }~dx

    \displaystyle   = \int \frac{d(x-1/x)}{(x-1/x)^2 + 9}

    \displaystyle   = \frac{1}{3} \tan^{-1}\left( \frac{x^2 - 1}{3x} \right)

    Therefore , the answer is

    \displaystyle   - \frac{1}{2\sqrt{5}} \tan^{-1}\left( \frac{x^2+1}{\sqrt{5} x } \right) +  \frac{5}{6} \tan^{-1}\left( \frac{x^2 - 1}{3x} \right) + C
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  8. #8
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    After Tonio's factorization, we can write the 2x^2+3 in terms of the sum of the two quadratic factors, i.e:
    We just observe that \displaystyle \left(x^2+\frac{7-3\sqrt{5}}{2}\right)+\left(x^2+\frac{7+3\sqrt{5}}{  2}\right) = 2x^2+7 then subtract 4.

    \displaystyle \begin{aligned}I & = \int\frac{2x^2+3}{x^4+7x^2+1}\;{dx} \\ & = \int \frac{2x^2+3}{\left(x^2+\frac{7-3\sqrt{5}}{2}\right)\left(x^2+\frac{7+3\sqrt{5}}{2  }\right)}\;{dx} \\ & = \int \frac{\left(x^2+\frac{7-3\sqrt{5}}{2}\right)+\left(x^2+\frac{7+3\sqrt{5}}{  2}\right)-4}{\left(x^2+\frac{7-3\sqrt{5}}{2}\right)\left(x^2+\frac{7+3\sqrt{5}}{2  }\right)}\;{dx} \\ & = \int \frac{1}{x^2+\frac{7+3\sqrt{5}}{2}}\;{dx}+ \int \frac{1}{x^2+\frac{7-3\sqrt{5}}{2}}\;{dx} -\int \frac{4}{\left(x^2+\frac{7-3\sqrt{5}}{2}\right)\left(x^2+\frac{7+3\sqrt{5}}{2  }\right)}\;{dx} \\ & = \frac{\sqrt{2}}{\sqrt{7+3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7+3\sqrt{5}}}\right  )+\frac{\sqrt{2}}{\sqrt{7-3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7-3\sqrt{5}}}\right)-I_{1}. \end{aligned}

    For I_{1}, we write 4 as in terms of the two quadratic factors, and we do it as follows -- let:

    \alpha\left(x^2+\frac{7+3\sqrt{5}}{2}\right)-\alpha\left(x^2+\frac{7-3\sqrt{5}}{2}\right)} = 4, then putting x = 0 gives you the value:

    \displaystyle \alpha  = \frac{4}{\frac{1}{2}(7+3\sqrt{5})-\frac{1}{2}(7-3\sqrt{5})} = \frac{4}{\frac{1}{2}(3\sqrt{5}+3\sqrt{5})} = \frac{4}{3\sqrt{5}}. Therefore, we have:

    \displaystyle \begin{aligned} I_{1} & = \int \frac{4}{\left(x^2+\frac{7-3\sqrt{5}}{2}\right)\left(x^2+\frac{7+3\sqrt{5}}{2  }\right)}\;{dx} \\ & = \int \frac{\frac{4}{3\sqrt{5}}\left(x^2+\frac{7+3\sqrt{  5}}{2}\right)-\frac{4}{3\sqrt{5}}\left(x^2+\frac{7-3\sqrt{5}}{2}\right)}{\left(x^2+\frac{7-3\sqrt{5}}{2}\right)\left(x^2+\frac{7+3\sqrt{5}}{2  }\right)}{dx} \\ & = \frac{4}{3\sqrt{5}}\int\frac{1}{x^2+\frac{7-3\sqrt{5}}{2}} \;{dx} - \frac{4}{3\sqrt{5}}\int\frac{1}{x^2+\frac{7+3\sqrt  {5}}{2}} \;{dx} \\ & = \frac{4\sqrt{2}}{3\sqrt{5}\sqrt{7-3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7-3\sqrt{5}}}\right)-\frac{4\sqrt{2}}{3\sqrt{5}\sqrt{7+3\sqrt{5}}}\tan^  {-1}\left(\frac{\sqrt{2}x}{\sqrt{7+3\sqrt{5}}}\right  )}. \end{aligned}

    Therefore, we have:

    \begin{aligned} I & = \frac{\sqrt{2}}{\sqrt{7+3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7+3\sqrt{5}}}\right  )+\frac{\sqrt{2}}{\sqrt{7-3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7-3\sqrt{5}}}\right) \\& -\frac{4\sqrt{2}}{3\sqrt{5}\sqrt{7-3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7-3\sqrt{5}}}\right)+\frac{4\sqrt{2}}{3\sqrt{5}\sqrt  {7+3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7+3\sqrt{5}}}\right  ) +k  \\& = \frac{\sqrt{2}}{\sqrt{7+3\sqrt{5}}}\left(1+\frac{4  }{3\sqrt{5}}\right)\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7+3\sqrt{5}}}\right  )  +\frac{\sqrt{2}}{\sqrt{7-3\sqrt{5}}}\left(1-\frac{4}{3\sqrt{5}}\right)\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7-3\sqrt{5}}}\right) \\& = \boxed{\frac{3\sqrt{2}\sqrt{5}+4\sqrt{2}}{3\sqrt{5  }\sqrt{7+3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7+3\sqrt{5}}}\right  )+\frac{3\sqrt{2}\sqrt{5}-4\sqrt{2}}{3\sqrt{5}\sqrt{7-3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7-3\sqrt{5}}}\right)+k}   \end{aligned}

    Isn't that just an absolute beauty? Beats all the sunsets one can see or imagine, haha! Now:

    \displaystyle \begin{aligned} \int_0^{\infty}\frac{2x^2+3}{x^4+7x^2+1} \; dx & = \frac{3\sqrt{2}\sqrt{5}+4\sqrt{2}}{3\sqrt{5}\sqrt{  7+3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7+3\sqrt{5}}}\right  )+\frac{3\sqrt{2}\sqrt{5}-4\sqrt{2}}{3\sqrt{5}\sqrt{7-3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7-3\sqrt{5}}}\right)\bigg|_{0}^{\infty} \\& = \lim_{x \to \infty}\bigg[\frac{3\sqrt{2}\sqrt{5}+4\sqrt{2}}{3\sqrt{5}\sqrt{  7+3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7+3\sqrt{5}}}\right  )+\frac{3\sqrt{2}\sqrt{5}-4\sqrt{2}}{3\sqrt{5}\sqrt{7-3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7-3\sqrt{5}}}\right)\bigg]  \\& = \frac{\pi}{2}\left(\frac{3\sqrt{2}\sqrt{5}+4\sqrt{  2}}{3\sqrt{5}\sqrt{7+3\sqrt{5}}}+\frac{3\sqrt{2}\s  qrt{5}-4\sqrt{2}}{3\sqrt{5}\sqrt{7-3\sqrt{5}}}\right) = \frac{\pi}{2}\left(\frac{5}{3}\right) = \frac{5\pi}{6}. \end{aligned}
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  9. #9
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    What's happening with my latex?

    Quote Originally Posted by simplependulum View Post
     \displaystyle  2x^2 + 3 = A(x^2 - 1) + B(x^2 + 1) where  A,B are constants , solving it we obtain  \displaystyle   A = -0.5 ~,~ B = 2.5
    I'm shocked by how you tweaked the integral just to make it admit those couple of substitutions! Nice!
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  10. #10
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by simplependulum View Post
    Write


     \displaystyle  2x^2 + 3 = A(x^2 - 1) + B(x^2 + 1) where  A,B are constants , solving it we obtain  \displaystyle   A = -0.5 ~,~ B = 2.5

    We have

     \displaystyle  \int \frac{2x^2 + 3}{x^4 + 7x^2 + 1}~dx

     \displaystyle  = -\frac{1}{2} \int \frac{x^2 - 1}{x^4 + 7x^2 + 1 }~dx  + \frac{5}{2} \int \frac{x^2 + 1}{x^4 + 7x^2 + 1}~dx

    For the first integral , consider

     \displaystyle   \int \frac{1 - 1/x^2}{x^2 + 7 + 1/x^2 }~dx

     \displaystyle  = \int \frac{d(x+1/x)}{(x+1/x)^2 + 5 }

    Sub. \displaystyle   u = x+1/x

    \displaystyle   = \int \frac{du}{u^2+5} = \frac{1}{\sqrt{5}} \tan^{-1}(u/\sqrt{5})

     \displaystyle  = \frac{1}{\sqrt{5}} \tan^{-1}\left( \frac{x^2+1}{\sqrt{5} x } \right)

    For the second integral , you may find it is a similar trick

    \displaystyle    \int \frac{1 + 1/x^2}{x^2 + 7 + 1/x^2 }~dx

    \displaystyle   = \int \frac{d(x-1/x)}{(x-1/x)^2 + 9}

    \displaystyle   = \frac{1}{3} \tan^{-1}\left( \frac{x^2 - 1}{3x} \right)

    Therefore , the answer is

    \displaystyle   - \frac{1}{2\sqrt{5}} \tan^{-1}\left( \frac{x^2+1}{\sqrt{5} x } \right) +  \frac{5}{6} \tan^{-1}\left( \frac{x^2 - 1}{3x} \right) + C
    wow thats cooool thanks for u all
    I like the way u think thats very cool I learned much from u
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  11. #11
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by simplependulum View Post
    Write

    For the first integral , consider

     \displaystyle   \int \frac{1 - 1/x^2}{x^2 + 7 + 1/x^2 }~dx

     \displaystyle  = \int \frac{d(x+1/x)}{(x+1/x)^2 + 5 }
    Edit- Oh wow. My bad. Too much to drink. Gotcha
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