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Math Help - Another bad integral?

  1. #1
    Newbie superphysics's Avatar
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    Another bad integral?

    This is another question that I'm having trouble solving... is it also bad?

     <br />
\int{\frac{x^2}{x^4+1}}dx<br />

    Thanks.
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  2. #2
    Eater of Worlds
    galactus's Avatar
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    It's not as bad as the other, but it ain't pretty neither.

    You could use a clever partial fraction decomp on it, may be one way.

    \frac{\sqrt{2}}{4}\int\frac{x}{x^{2}-\sqrt{2}x+1}dx-\frac{\sqrt{2}}{4}\int\frac{x}{x^{2}+\sqrt{2}x+1}d  x
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  3. #3
    Newbie superphysics's Avatar
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    Thanks. It ain't pretty, as you say, but it works after a loooong manipulation.
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  4. #4
    Senior Member DivideBy0's Avatar
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    Quote Originally Posted by superphysics View Post
    This is another question that I'm having trouble solving... is it also bad?

     <br />
\int{\frac{x^2}{x^4+1}}dx<br />

    Thanks.
    I really should find something better to do with my time. Well, at least the LaTeX bashing was fun and it looks vaguely cool.

    I actually believe that by some really farfetched coincidence my answer is right.

    The denominator can factor as:

    \int \frac{x^2}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}\,dx

    You can then use partial fractions lol

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

    =\frac{\sqrt{2}}{4}\int \frac{x}{x^2-\sqrt{2}x+1}\,dx-\frac{\sqrt{2}}{4}\int \frac{x}{x^2+\sqrt{2}x+1}\,dx


    =\frac{\sqrt{2}}{4}\int \frac{x}{(x-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}\,dx-\frac{\sqrt{2}}{4}\int \frac{x}{(x+\frac{\sqrt{2}}{2})^2+\frac{1}{2}}\,dx

    Using \int \frac{z}{a^2+z^2}\,dz=\frac{1}{2}\ln(a^2+z^2)+C and \int \frac{\,dz}{a^2+z^2} = \frac{1}{a}\tan^{-1}{\frac{z}{a}}+C, let u = x+\frac{\sqrt{2}}{2} and v = x-\frac{\sqrt{2}}{2}, then the integral becomes

    \frac{\sqrt{2}}{4}\int \frac{u-\frac{\sqrt{2}}{2}}{u^2+\frac{1}{2}}\,du-\frac{\sqrt{2}}{4}\int \frac{v+\frac{\sqrt{2}}{2}}{v^2+\frac{1}{2}}\,dv

    =\left(\frac{\sqrt{2}}{4}\int \frac{u}{u^2+\frac{1}{2}}\,du - \frac{\sqrt{2}}{4}\int \frac{\frac{\sqrt{2}}{2}}{u^2+\frac{1}{2}}\,du\rig  ht)-\left(\frac{\sqrt{2}}{4} \int \frac{v}{v^2+\frac{1}{2}}\,dv+ \frac{\sqrt{2}}{4}\int \frac{\frac{\sqrt{2}}{2}}{v^2+\frac{1}{2}}\,dv\rig  ht)

    =\left(\frac{\sqrt{2}}{4}\int \frac{u}{u^2+\frac{1}{2}}\,du - \frac{1}{4}\int \frac{1}{u^2+\frac{1}{2}}\,du\right)-\left(\frac{\sqrt{2}}{4} \int \frac{v}{v^2+\frac{1}{2}}\,dv+ \frac{1}{4}\int \frac{1}{v^2+\frac{1}{2}}\,dv\right)

    =\frac{\sqrt{2}}{4}\left(\frac{1}{2}\ln(u^2+\frac{  1}{4})\right)-\frac{1}{4}\left(\frac{1}{\frac{\sqrt{2}}{2}}\tan^  {-1}\frac{u}{\frac{\sqrt{2}}{2}}\right)-\frac{\sqrt{2}}{4}\left(\frac{1}{2}\ln(v^2+\frac{1  }{4})\right)-\frac{1}{4}\left(\frac{1}{\frac{\sqrt{2}}{2}}\tan^  {-1}\frac{v}{\frac{\sqrt{2}}{2}}\right)

    =\frac{\sqrt{2}}{8}\ln(u^2+\frac{1}{4})-\frac{1}{2\sqrt{2}}\tan^{-1}\frac{2u}{\sqrt{2}}-\frac{\sqrt{2}}{8}\ln(v^2+\frac{1}{4})-\frac{1}{2\sqrt{2}}\tan^{-1}\frac{2v}{\sqrt{2}}

    =\frac{\sqrt{2}}{8}\ln\left((x+\frac{\sqrt{2}}{2})  ^2+\frac{1}{4}\right)-\frac{1}{2\sqrt{2}}\tan^{-1}\frac{2(x+\frac{\sqrt{2}}{2})}{\sqrt{2}}
    \ \ -\frac{\sqrt{2}}{8}\ln\left((x-\frac{\sqrt{2}}{2})^2+\frac{1}{4}\right)-\frac{1}{2\sqrt{2}}\tan^{-1}\frac{2(x-\frac{\sqrt{2}}{2})}{\sqrt{2}}


    =\frac{\sqrt{2}}{8}\ln\left((x+\frac{\sqrt{2}}{2})  ^2+\frac{1}{4}\right)-\frac{1}{2\sqrt{2}}\tan^{-1}{(\sqrt{2}x+1)}

    \ \ -\frac{\sqrt{2}}{8}\ln\left((x-\frac{\sqrt{2}}{2})^2+\frac{1}{4}\right)-\frac{1}{2\sqrt{2}}\tan^{-1}{(\sqrt{2}x-1)}

    +\ C

    Now I'm going to collapse and sleep for a while.
    Last edited by DivideBy0; December 15th 2007 at 10:59 PM. Reason: fixed typo pointed out by mr. fantastic
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  5. #5
    Math Engineering Student
    Krizalid's Avatar
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    Workin' with partial fractions it's just a mess, but possible.

    Here's a specific calculation in this case.

    Quote Originally Posted by superphysics View Post
     <br />
\int{\frac{x^2}{x^4+1}}dx<br />
    \int {\frac{{x^2 }}<br />
{{x^4 + 1}}\,dx} = \frac{1}<br />
{2}\int {\frac{{\left( {x^2 + 1} \right) + \left( {x^2 - 1} \right)}}<br />
{{x^4 + 1}}\,dx} = \frac{1}<br />
{2}\left[ {\int {\frac{{1 + \dfrac{1}<br />
{{x^2 }}}}<br />
{{x^2 + \dfrac{1}<br />
{{x^2 }}}}\,dx} + \int {\frac{{1 - \dfrac{1}<br />
{{x^2 }}}}<br />
{{x^2 + \dfrac{1}<br />
{{x^2 }}}}\,dx} } \right].

    Now

    \int {\frac{{1 + \dfrac{1}<br />
{{x^2 }}}}<br />
{{x^2 + \dfrac{1}<br />
{{x^2 }}}}\,dx} = \int {\frac{{\left( {x - \dfrac{1}<br />
{x}} \right)'}}<br />
{{\left( {x - \dfrac{1}<br />
{x}} \right)^2 + 2}}\,dx} = \frac{1}<br />
{{\sqrt 2 }}\arctan \frac{{x^2 - 1}}<br />
{{\sqrt 2 x}}.

    \int {\frac{{1 - \dfrac{1}<br />
{{x^2 }}}}<br />
{{x^2 + \dfrac{1}<br />
{{x^2 }}}}\,dx} = \int {\frac{{\left( {x + \dfrac{1}<br />
{x}} \right)'}}<br />
{{\left( {x + \dfrac{1}<br />
{x}} \right)^2 - 2}}\,dx} = \frac{1}<br />
{{2\sqrt 2 }}\ln \left| {\frac{{x^2 - \sqrt 2 x + 1}}<br />
{{x^2 + \sqrt 2 x + 1}}} \right|.

    Addin' the constant of integration we happily get

    \int {\frac{{x^2 }}<br />
{{x^4 + 1}}\,dx} = \frac{1}<br />
{2}\left[ {\frac{1}<br />
{{\sqrt 2 }}\arctan \frac{{x^2 - 1}}<br />
{{\sqrt 2 x}} + \frac{1}<br />
{{2\sqrt 2 }}\ln \left| {\frac{{x^2 - \sqrt 2 x + 1}}<br />
{{x^2 + \sqrt 2 x + 1}}} \right|} \right] + k.
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  6. #6
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by superphysics View Post
    This is another question that I'm having trouble solving... is it also bad?

     <br />
\int{\frac{x^2}{x^4+1}}dx<br />

    Thanks.
    Observation: You'll get (a multiple of) this integral in the first instance when integrating <br />
\int \sqrt{\tan t} \, dt \, .<br />
    Last edited by mr fantastic; December 15th 2007 at 08:59 PM. Reason: Wrong latex command
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
    \int {\frac{{x^2 }}<br />
{{x^4 + 1}}\,dx} = \frac{1}<br />
{2}\left[ {\frac{1}<br />
{{\sqrt 2 }}\arctan \frac{{x^2 - 1}}<br />
{{\sqrt 2 x}} + \frac{1}<br />
{{2\sqrt 2 }}\ln \left| {\frac{{x^2 - \sqrt 2 x + 1}}<br />
{{x^2 + \sqrt 2 x + 1}}} \right|} \right] + k.
    Quote Originally Posted by DivideBy0 View Post
    =\frac{\sqrt{2}}{8}\ln\left((x+\frac{\sqrt{2}}{2})  ^2+\frac{1}{4}\right)-\frac{1}{2\sqrt{2}}\tan^{-1}{(2x+1)}

    \ \ -\frac{\sqrt{2}}{8}\ln\left((x-\frac{\sqrt{2}}{2})^2+\frac{1}{4}\right)-\frac{1}{2\sqrt{2}}\tan^{-1}{(2x-1)}

    +\ C
    I find it fascinating that the two of these are different only by a constant.

    One question: Krizalid's solution will not work if the domain of integration includes 0 (or does it?). Why does DivideBy0's solution work for such a region?

    -Dan
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  8. #8
    Flow Master
    mr fantastic's Avatar
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    Not >that< fascinating

    Quote Originally Posted by topsquark View Post
    I find it fascinating that the two of these are different only by a constant. [snip]
    -Dan
    Well, it's not that fascinating really. The two log terms are easily combined and the result simplified to almost get Krizalid's log term. The two arctan terms are less but not too less easily combined and simplified to almost get Krizalid's arctan term.

    The reason for the "almost" is that there are a couple of typos(?) in DivideBy0's answer. Eg 2x + 1 rather than \sqrt{2}x + 1 etx. .....
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  9. #9
    Flow Master
    mr fantastic's Avatar
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    It does work

    Quote Originally Posted by topsquark View Post
    [snip]

    One question: Krizalid's solution will not work if the domain of integration includes 0 (or does it?). Why does DivideBy0's solution work for such a region?

    -Dan
    Both work:

    arctan(oo) is undefined but could be thought of as being equal to -pi/2 (note that -tan pi/2 is undefined).

    And obviously -pi/4 - pi/4 = -pi/2 .......
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