Question

\[\int\frac{1}{1+x^{4}}dx\]

Answer 1

Without integration by partial fractions.

Answer 2

can i give it now? :S

Answer 3

Go ahead. :P

Answer 4

ill wait some more :p

Answer 5

naah, noow; just see what you got :P

Answer 6

is it trigonometric substitution?

Answer 7

You can manipulate it; not sure if trig-sub would work. I am not sure if \(x^2=tan\theta\) would work.

@lgbasallote: Can tell us his method. :P

Answer 8

ill let @Mimi_x3 express his/her methods first

Answer 9

lol, who is Mimi_x3?

Answer 10

No one remembered me I see. :\

Still. Divide both numerator and denominator by x^2

Answer 11

trig sub get messy...:(

Answer 12

yeah, it does; but the manipulation gets messier i suppose :/

Answer 13

\[\int \frac{\frac1{x^2}}{\frac1{x^2} + x^2}.dx\]
\[=\int \frac{\frac1{x^2}}{(\frac1{x} +x)^2-2}.dx\]

Now, let "x + 1/x" be t
the, (-1/(x^2)).dx = dt

So,
\[\int \frac{-dt}{t^2 -2}\]

INTEGRATE THIS!!!!

Answer 14

The loving integrand formula for this one is:\[\int\frac{dx}{x^2 - a^2}= \frac{1}{2a}ln(\frac{x-a}{x+a}) + c\]

Answer 15

Nice manipulation; i havent though of that.
\[\frac{1}{2} \int\limits\frac{2}{1+x^{2}} dx => \frac{1}{2} \int\limits\frac{(x^{2}+1)-(x^{2}-1)}{1+x^{4}} \]
this is another way (:

Answer 16

Yeah, this is another way, but then you have two integrals, so a wee bit messier, but nonetheless , a good one :)

Answer 17

I made a typo. xD
\[\frac{1}{2} \int\limits\frac{2}{1+x^{4}} dx\]

Answer 18

another way
\[ 1 + x^4 = (1 + ix^2)(1 - ix^2)\]

Answer 19

lol, complex numbers..

Answer 20

if you are comfortable with complex numbers ...

Answer 21

yeah, abit harder i suppose..

Answer 22

Hold on....
Let t = x + 1/x
dt = 1 - 1/x^2 dx .... Isn't it?
(Sorry, I haven't done differentiation and integration for long..)

Answer 23

I mean ... if dt = 1 - 1/x^2 dx is true, then (-1/(x^2)).dx = dt is not correct...
Then, we cannot get \(\int \frac{-dt}{t^2 -2}\)

Answer 24

@Callisto you are right, I blooped big time. :/ Thanks...
Actually i should have done it this way:

\[=\int \frac{\frac1{x^2}}{(x -\frac1{x} )^2+2}.dx\]

Now it should work out like this:
Let x - 1/x be 't'
Then
(1/(x^2)).dx = dt

So,
\[\int \frac{dt}{t^2 + 2}dt\]

And this one's even easier.
\[\int \frac{dx}{x^2+a^2} = \frac 1 a \tan^{-1}\frac x a + c\]

Answer 25

No... I meant
Let x - 1/x be 't'
Then
(1/(x^2)).dx = dt <- but isn't it dt = 1+ 1/x^2 dx ?

Answer 26

I AM SO SO SOOOOOOOOOOOO STUPID!!!! Take back those virtual accolades, i don't deserve them! :(

Answer 27

lol, forget about this integral :)

Answer 28

No... The method is good. But the question is still unsolved..

Answer 29

Don't worry about it (:

I already gave you a method that would work.

Answer 30

Okay, i seem to have got his now.

\[=\int \frac{\frac1{x^2}}{x^2 +\frac1{x^2} }.dx =\frac 1 2 ( \int \frac{1+\frac1{x^2}}{(x -\frac1{x} )^2+2}.dx - \int \frac{1 - \frac1{ x^2} }{(x +\frac1{x} )^2-2}.dx) \]


Hmm should work now.

Answer 31

*this

lool.

Answer 32

I hope I didn't do anything stupid again. -_-

Answer 33

Okay let's work this thing out, just to be sure:
In the first part, let
\[x - \frac 1 x = t\]\[or, (1 + \frac 1 {x^2})dx = dt\]

In the second part,
\[Let~~ x +\frac 1 x = m\]\[So,~~ (1 - \frac 1 {x^2})dx = dm\]

So, what we have is:
\[\frac 1 2 (\int \frac{dt}{t^2 + 2} - \int \frac{dt}{m^2 - 2})\]
\[=\frac 1 2 ( \frac 1 {\sqrt 2}tan^{-1}(\frac t {\sqrt 2}) - \frac 1 {2\sqrt 2}ln(\frac{m-\sqrt2}{m +\sqrt2}) + k)\]

Now substitute t = x - 1/x and m = x + 1/x, and enjoy!!

Answer 34

@Callisto: Another method.
\[\frac{1}{2} \int\limits\frac{2}{1+x^{4}} dx => \frac{1}{2} \int\limits\frac{(x^{2}+1)-(x^{2}-1)}{1+x^{4}} =>\]
\[\frac{1}{2} \int\limits\frac{x^{2}+1}{1+x^{4}} -\frac{x^{2}-1}{1+x^{4}} \]
I think that you're able to do it now.