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Loser66

  • one year ago

\(\int \dfrac{1}{1+x^4}dx\) Appreciate any tip.

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  1. SolomonZelman
    • one year ago
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    are you sure that is the question ?

  2. SolomonZelman
    • one year ago
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    http://www.wolframalpha.com/input/?i=integral+of+1%2F%281%2Bx%5E4%29

  3. Loser66
    • one year ago
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    yes!! @SolomonZelman

  4. SolomonZelman
    • one year ago
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    maybe that is a minus on the bottom

  5. SolomonZelman
    • one year ago
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    a long long one.

  6. Loser66
    • one year ago
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    @freckles I knew how to solve the original one, just wonder how long does it take if we go the traditional way.

  7. SolomonZelman
    • one year ago
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    original one ?

  8. Loser66
    • one year ago
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    yes, it asked me to find the graph of solution of \(\dfrac{dy}{dx}= 1+y^4\) and 4 options. I know how to solve it.

  9. SolomonZelman
    • one year ago
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    I see, so you used a separation of variables, and got that integral.

  10. Loser66
    • one year ago
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    just wonder, if I solve that differential equation by traditional way, how long it will take. :)

  11. Loser66
    • one year ago
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    I have student. Have to leave now.

  12. SolomonZelman
    • one year ago
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    i would prefer stepsize for a particular point on the function, but this is just too much work I think.

  13. SolomonZelman
    • one year ago
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    or, getting a direction field from mathematica, .. idk.

  14. SolomonZelman
    • one year ago
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    oh, a power series for the integral

  15. SolomonZelman
    • one year ago
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    \(\large\color{slate}{\displaystyle \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n}\) \(\large\color{slate}{\displaystyle \frac{1}{1+x^4}=\sum_{n=0}^{\infty}(-1)^nx^{4n}}\) \(\large\color{slate}{\displaystyle \int\frac{1}{1+x^4}dx=\sum_{n=0}^{\infty}\frac{(-1)^nx^{4n+1}}{4n+1}}\)

  16. sidsiddhartha
    • one year ago
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    why dont u people try residue

  17. sidsiddhartha
    • one year ago
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    consider \[f(z)=\frac{ 1 }{ 1+z^4 }\\\] poles can be calculated now \[1+z^4=0\\z^4=-1\\z^4=(\cos \pi +i \sin \pi)=\cos(2n+1)\pi +i \sin(2n+1)\pi\\z=[\cos \frac{ 2n+1 }{ 4 }\pi+\sin \frac{ 2n+1 }{ 4 }\pi]\] we will get 4 poles at n=0,1,2,3 then we can proceed with residue

  18. freckles
    • one year ago
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    \[\frac{1}{x^4+1}=\frac{Ax+B}{x^2+\sqrt{2}x+1}+\frac{Cx+D}{x^2-\sqrt{2}x+1} \\ 1=x^3(A+C)+x^2(-\sqrt{2}A+B+\sqrt{2}C+D)+x(-\sqrt{2}B+A+C+\sqrt{2}D) \\ +(B+D) \\ A=-C \\ 2 \sqrt{2}C+B+(1-B)=0 \\ 2 \sqrt{2}C=-1 \\ C=\frac{-1}{2 \sqrt{2}} \\ A =\frac{1}{2 \sqrt{2}} \\ -\sqrt{2}B+\sqrt{2}(1-B)=0 \\ - 2 \sqrt{2} B=-\sqrt{2} \\ 2B=1 \\ B=\frac{1}{2} \\ \int\limits \frac{1}{x^4+1} dx=\int\limits \frac{\frac{1}{2 \sqrt{2}}x+\frac{1}{2}}{x^2+\sqrt{2}x+1} dx+\int\limits \frac{-\frac{1}{2 \sqrt{2}}x+\frac{1}{2}}{x^2-\sqrt{2}x+1} dx\] \[u=x^2+\sqrt{2}x+1 \\ du=(2x+\sqrt{2}) dx \\ \\ \text{ hmm... well } \frac{1}{2 \sqrt{2}}=\frac{\sqrt{2}}{2(2)}=\frac{\sqrt{2}}{4} \\ \sqrt{2} du=(2 \sqrt{2} x +2) dx \\ \frac{\sqrt{2}}{8} du=(\frac{\sqrt{2}}{4} x+\frac{1}{4}) dx \\ \text{ and } v=x^2- \sqrt{2}x+1 \\ \text{ gives } \frac{\sqrt{2}}{8}dv=(\frac{\sqrt{2}}{4}x-\frac{1}{4}) dx \\ \int\limits \frac{1}{x^4+1} dx= \int\limits \frac{\frac{\sqrt{2}}{8} du}{u}+\int\limits \frac{\frac{-\sqrt{2}}{8}}{v} dv+\int\limits \frac{\frac{1}{4}}{x^2+\sqrt{2}x+1} dx+\int\limits \frac{\frac{1}{4}}{x^2-\sqrt{2}x+1} dx\] \[\int\limits \frac{1}{x^4+1} dx \\ =\frac{\sqrt{2}}{8} \ln|x^2+\sqrt{2}x+1|-\frac{\sqrt{2}}{8} \ln|x^2-\sqrt{2}x+1|+\frac{1}{4} \int\limits \frac{dx}{(x+\frac{\sqrt{2}}{2})^2+\frac{1}{2}} dx \\ +\frac{1}{4} \int\limits \frac{dx}{(x-\frac{\sqrt{2}}{2})^2+\frac{1}{2}} \\ =\frac{\sqrt{2}}{8}[\ln|x^2-\sqrt{2}x+1|+\ln|x^2-\sqrt{2}x+1|]+\frac{\sqrt{2}}{4} \arctan(\sqrt{2}x+1) \\ +\frac{ \sqrt{2}}{4} \arctan(\sqrt{2}x-1)+C\]

  19. anonymous
    • one year ago
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    Another path you might consider: \[\begin{align*}\frac{1}{x^4+1}&=\frac{1}{2}\left(\frac{x^2+1}{x^4+1}-\frac{x^2-1}{x^4+1}\right)\\\\ &=\frac{1}{2}\left(\frac{1+\dfrac{1}{x^2}}{x^2+\dfrac{1}{x^2}}-\frac{1-\dfrac{1}{x^2}}{x^2+\dfrac{1}{x^2}}\right)\\\\ &=\frac{1}{2}\left(\frac{1+\dfrac{1}{x^2}}{\left(x-\dfrac{1}{x}\right)^2+2}-\frac{1-\dfrac{1}{x^2}}{\left(x+\dfrac{1}{x}\right)^2-2}\right) \end{align*}\] Setting \(u_1=x-\dfrac{1}{x}\) and \(u_2=x+\dfrac{1}{x}\) gives \(du_1=\left(1+\dfrac{1}{x}^2\right)\,dx\) and \(du_2=\left(1-\dfrac{1}{x}^2\right)\,dx\), so you have the integrals \[\frac{1}{2}\left(\int\frac{du_1}{{u_1}^2+2}-\int\frac{du_2}{({u_2}^2-2)}\right)\]

  20. freckles
    • one year ago
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    actually wrote the bottom like that but couldn't figure out any tricks to make it work nice job sith

  21. Loser66
    • one year ago
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    @SithsAndGiggles It is a very elegant solution. Do we have to see the problem before to quickly get that solution? If we don't , how to get it?

  22. Loser66
    • one year ago
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    @freckles , this is the problem

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  23. Loser66
    • one year ago
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    In 2 minutes, we can't solve the problem logically like what we did above, right? My logic is A is a correct answer because if y = 0, y' =1 y =1, y' = 2 y = 2, y'= 17 y =3, y'= 82 y=4, y'= 257 Hence, look at the graph A, the slope of the tangent lines of the curve "seem" satisfy what I did. The graph is almost a vertical line after ... around 3 or 4

  24. Loser66
    • one year ago
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    if B is the correct one, when y =0, the slope doesn't look like 1 C can't be the correct answer because the slope of the curve is near 0 when x gets bigger and y stays D can't not be the correct answer with the same reason with C when y approaches 0 y' never =0 hence, E can't be the correct one

  25. Loser66
    • one year ago
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    As I stated above, in 2 minutes, I have to guess the answer but I am not sure whether my logic is valid or not. Please, give me your opinion.

  26. freckles
    • one year ago
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    we definitely have y'>0 which means y is increasing function so we can totally throw out choice E y' is also never 0 so we can throw choice B now y' is an even function so y is odd function so we are looking at choice A or choice C as y gets large y' get large and C is not showing y' is getting large C actually appears to have a horizontal asymptote at y=c and y=-c where c is some positive number so in ruling out all the other choices A sounds great

  27. Loser66
    • one year ago
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    Got you. Thanks a lot.

  28. anonymous
    • one year ago
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    @Loser66 the partial fraction approach is probably the easier route. First, substitute \(t=x^2\) (just to make the next steps clearer): \[x^4+1=t^2+1\] Now let's complete the square by introducing a linear term: \[\begin{align*} t^2+1&=t^2+2t+1-2t\\ &=(t+1)^2-2t\\ &=(x^2+1)^2-2x^2&\text{since }t=x^2\\ &=((x^2+1)-\sqrt2 x)((x^2+1)+\sqrt2 x)&\text{difference of squares}\\ x^4+1&=(x^2-\sqrt 2x+1)(x^2+\sqrt2x+1) \end{align*}\]

  29. Loser66
    • one year ago
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    Yeeeeeeees!! It is perfect. Thank you so much.

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