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Loser66
 one year ago
\(\int \dfrac{1}{1+x^4}dx\)
Appreciate any tip.
Loser66
 one year ago
\(\int \dfrac{1}{1+x^4}dx\) Appreciate any tip.

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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0are you sure that is the question ?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=integral+of+1%2F%281%2Bx%5E4%29

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0maybe that is a minus on the bottom

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0a long long one.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@freckles I knew how to solve the original one, just wonder how long does it take if we go the traditional way.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0yes, 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.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0I see, so you used a separation of variables, and got that integral.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0just wonder, if I solve that differential equation by traditional way, how long it will take. :)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I have student. Have to leave now.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0i would prefer stepsize for a particular point on the function, but this is just too much work I think.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0or, getting a direction field from mathematica, .. idk.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0oh, a power series for the integral

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\color{slate}{\displaystyle \frac{1}{1+x}=\sum_{n=0}^{\infty}(1)^nx^n}\) \(\large\color{slate}{\displaystyle \frac{1}{1+x^4}=\sum_{n=0}^{\infty}(1)^nx^{4n}}\) \(\large\color{slate}{\displaystyle \int\frac{1}{1+x^4}dx=\sum_{n=0}^{\infty}\frac{(1)^nx^{4n+1}}{4n+1}}\)

sidsiddhartha
 one year ago
Best ResponseYou've already chosen the best response.0why dont u people try residue

sidsiddhartha
 one year ago
Best ResponseYou've already chosen the best response.0consider \[f(z)=\frac{ 1 }{ 1+z^4 }\\\] 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{1}{x^4+1}=\frac{Ax+B}{x^2+\sqrt{2}x+1}+\frac{Cx+D}{x^2\sqrt{2}x+1} \\ 1=x^3(A+C)+x^2(\sqrt{2}A+B+\sqrt{2}C+D)+x(\sqrt{2}B+A+C+\sqrt{2}D) \\ +(B+D) \\ A=C \\ 2 \sqrt{2}C+B+(1B)=0 \\ 2 \sqrt{2}C=1 \\ C=\frac{1}{2 \sqrt{2}} \\ A =\frac{1}{2 \sqrt{2}} \\ \sqrt{2}B+\sqrt{2}(1B)=0 \\  2 \sqrt{2} B=\sqrt{2} \\ 2B=1 \\ B=\frac{1}{2} \\ \int\limits \frac{1}{x^4+1} dx=\int\limits \frac{\frac{1}{2 \sqrt{2}}x+\frac{1}{2}}{x^2+\sqrt{2}x+1} dx+\int\limits \frac{\frac{1}{2 \sqrt{2}}x+\frac{1}{2}}{x^2\sqrt{2}x+1} dx\] \[u=x^2+\sqrt{2}x+1 \\ du=(2x+\sqrt{2}) dx \\ \\ \text{ hmm... well } \frac{1}{2 \sqrt{2}}=\frac{\sqrt{2}}{2(2)}=\frac{\sqrt{2}}{4} \\ \sqrt{2} du=(2 \sqrt{2} x +2) dx \\ \frac{\sqrt{2}}{8} du=(\frac{\sqrt{2}}{4} x+\frac{1}{4}) dx \\ \text{ and } v=x^2 \sqrt{2}x+1 \\ \text{ gives } \frac{\sqrt{2}}{8}dv=(\frac{\sqrt{2}}{4}x\frac{1}{4}) dx \\ \int\limits \frac{1}{x^4+1} dx= \int\limits \frac{\frac{\sqrt{2}}{8} du}{u}+\int\limits \frac{\frac{\sqrt{2}}{8}}{v} dv+\int\limits \frac{\frac{1}{4}}{x^2+\sqrt{2}x+1} dx+\int\limits \frac{\frac{1}{4}}{x^2\sqrt{2}x+1} dx\] \[\int\limits \frac{1}{x^4+1} dx \\ =\frac{\sqrt{2}}{8} \lnx^2+\sqrt{2}x+1\frac{\sqrt{2}}{8} \lnx^2\sqrt{2}x+1+\frac{1}{4} \int\limits \frac{dx}{(x+\frac{\sqrt{2}}{2})^2+\frac{1}{2}} dx \\ +\frac{1}{4} \int\limits \frac{dx}{(x\frac{\sqrt{2}}{2})^2+\frac{1}{2}} \\ =\frac{\sqrt{2}}{8}[\lnx^2\sqrt{2}x+1+\lnx^2\sqrt{2}x+1]+\frac{\sqrt{2}}{4} \arctan(\sqrt{2}x+1) \\ +\frac{ \sqrt{2}}{4} \arctan(\sqrt{2}x1)+C\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Another path you might consider: \[\begin{align*}\frac{1}{x^4+1}&=\frac{1}{2}\left(\frac{x^2+1}{x^4+1}\frac{x^21}{x^4+1}\right)\\\\ &=\frac{1}{2}\left(\frac{1+\dfrac{1}{x^2}}{x^2+\dfrac{1}{x^2}}\frac{1\dfrac{1}{x^2}}{x^2+\dfrac{1}{x^2}}\right)\\\\ &=\frac{1}{2}\left(\frac{1+\dfrac{1}{x^2}}{\left(x\dfrac{1}{x}\right)^2+2}\frac{1\dfrac{1}{x^2}}{\left(x+\dfrac{1}{x}\right)^22}\right) \end{align*}\] 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 \[\frac{1}{2}\left(\int\frac{du_1}{{u_1}^2+2}\int\frac{du_2}{({u_2}^22)}\right)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1actually wrote the bottom like that but couldn't figure out any tricks to make it work nice job sith

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@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?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@freckles , this is the problem

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0In 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

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0if 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

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0As 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.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1we 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

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Got you. Thanks a lot.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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: \[\begin{align*} t^2+1&=t^2+2t+12t\\ &=(t+1)^22t\\ &=(x^2+1)^22x^2&\text{since }t=x^2\\ &=((x^2+1)\sqrt2 x)((x^2+1)+\sqrt2 x)&\text{difference of squares}\\ x^4+1&=(x^2\sqrt 2x+1)(x^2+\sqrt2x+1) \end{align*}\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Yeeeeeeees!! It is perfect. Thank you so much.
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