Question

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

Answer 1

are you sure that is the question ?

Answer 2

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

Answer 3

yes!! @SolomonZelman

Answer 4

maybe that is a minus on the bottom

Answer 5

a long long one.

Answer 6

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

Answer 7

original one ?

Answer 8

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.

Answer 9

I see, so you used a separation of variables, and got that integral.

Answer 10

just wonder, if I solve that differential equation by traditional way, how long it will take. :)

Answer 11

I have student. Have to leave now.

Answer 12

i would prefer stepsize for a particular point on the function, but this is just too much work I think.

Answer 13

or, getting a direction field from mathematica, .. idk.

Answer 14

oh, a power series for the integral

Answer 15

\(\large\color{slate}{\displaystyle \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n}\)

\(\large\color{slate}{\displaystyle \frac{1}{1+x^4}=\sum_{n=0}^{\infty}(-1)^nx^{4n}}\)

\(\large\color{slate}{\displaystyle \int\frac{1}{1+x^4}dx=\sum_{n=0}^{\infty}\frac{(-1)^nx^{4n+1}}{4n+1}}\)

Answer 16

why dont u people try residue

Answer 17

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

Answer 18

\[\frac{1}{x^4+1}=\frac{Ax+B}{x^2+\sqrt{2}x+1}+\frac{Cx+D}{x^2-\sqrt{2}x+1} \\ 1=x^3(A+C)+x^2(-\sqrt{2}A+B+\sqrt{2}C+D)+x(-\sqrt{2}B+A+C+\sqrt{2}D) \\ +(B+D) \\ A=-C \\ 2 \sqrt{2}C+B+(1-B)=0 \\ 2 \sqrt{2}C=-1 \\ C=\frac{-1}{2 \sqrt{2}} \\ A =\frac{1}{2 \sqrt{2}} \\ -\sqrt{2}B+\sqrt{2}(1-B)=0 \\ - 2 \sqrt{2} B=-\sqrt{2} \\ 2B=1 \\ B=\frac{1}{2} \\ \int\limits \frac{1}{x^4+1} dx=\int\limits \frac{\frac{1}{2 \sqrt{2}}x+\frac{1}{2}}{x^2+\sqrt{2}x+1} dx+\int\limits \frac{-\frac{1}{2 \sqrt{2}}x+\frac{1}{2}}{x^2-\sqrt{2}x+1} dx\]

\[u=x^2+\sqrt{2}x+1 \\ du=(2x+\sqrt{2}) dx \\ \\ \text{ hmm... well } \frac{1}{2 \sqrt{2}}=\frac{\sqrt{2}}{2(2)}=\frac{\sqrt{2}}{4} \\ \sqrt{2} du=(2 \sqrt{2} x +2) dx \\ \frac{\sqrt{2}}{8} du=(\frac{\sqrt{2}}{4} x+\frac{1}{4}) dx \\ \text{ and } v=x^2-
\sqrt{2}x+1 \\ \text{ gives } \frac{\sqrt{2}}{8}dv=(\frac{\sqrt{2}}{4}x-\frac{1}{4}) dx \\ \int\limits \frac{1}{x^4+1} dx= \int\limits \frac{\frac{\sqrt{2}}{8} du}{u}+\int\limits \frac{\frac{-\sqrt{2}}{8}}{v} dv+\int\limits \frac{\frac{1}{4}}{x^2+\sqrt{2}x+1} dx+\int\limits \frac{\frac{1}{4}}{x^2-\sqrt{2}x+1} dx\]

\[\int\limits \frac{1}{x^4+1} dx \\ =\frac{\sqrt{2}}{8} \ln|x^2+\sqrt{2}x+1|-\frac{\sqrt{2}}{8} \ln|x^2-\sqrt{2}x+1|+\frac{1}{4} \int\limits \frac{dx}{(x+\frac{\sqrt{2}}{2})^2+\frac{1}{2}} dx \\ +\frac{1}{4} \int\limits \frac{dx}{(x-\frac{\sqrt{2}}{2})^2+\frac{1}{2}} \\ =\frac{\sqrt{2}}{8}[\ln|x^2-\sqrt{2}x+1|+\ln|x^2-\sqrt{2}x+1|]+\frac{\sqrt{2}}{4} \arctan(\sqrt{2}x+1) \\ +\frac{ \sqrt{2}}{4} \arctan(\sqrt{2}x-1)+C\]

Answer 19

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
\[\frac{1}{2}\left(\int\frac{du_1}{{u_1}^2+2}-\int\frac{du_2}{({u_2}^2-2)}\right)\]

Answer 20

actually wrote the bottom like that
but couldn't figure out any tricks to make it work
nice job sith

Answer 21

@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?

Answer 22

@freckles , this is the problem

Answer 23

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

Answer 24

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

Answer 25

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.

Answer 26

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

Answer 27

Got you. Thanks a lot.

Answer 28

@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:
\[\begin{align*}
t^2+1&=t^2+2t+1-2t\\
&=(t+1)^2-2t\\
&=(x^2+1)^2-2x^2&\text{since }t=x^2\\
&=((x^2+1)-\sqrt2 x)((x^2+1)+\sqrt2 x)&\text{difference of squares}\\
x^4+1&=(x^2-\sqrt 2x+1)(x^2+\sqrt2x+1)
\end{align*}\]

Answer 29

Yeeeeeeees!! It is perfect. Thank you so much.