Question

what is the integral of (x)/(1+x^4)

Answer 1

PLEASE HELP.

Answer 2

(x)/(1+x^4)
i'm pretty sure that integrates into a trig function, i'm going to pen and paper it dude, brb k?

Answer 3

yep, that's right
integral of 1/x^2 + 1 = arctan (x) ... +c
\[\int\limits_{}^{} \frac{ 1 }{ x^2 + 1 } = \tan^{-1} (x) + c\]

Answer 4

so for your q:
\[\int\limits_{}^{} \frac {x}{x^4 +1} dx\]
so sub in x^2 as u
and make du = xdx
so
\[\int\limits_{}^{} \frac {1}{u^2 +1} du\]
\[\frac 12 tan^{-1} (u) +c \] and u = x^2
so...
\[\frac 12 tan^{-1} (x^2) +c \]

Answer 5

One mistake...
u = x^2
du = 2xdx

Answer 6

\[\int \frac {x}{x^4 +1} dx\]\[=\frac{1}{2}\int \frac {1}{u^2 +1} du\]

Answer 7

...why 2x...?
top line is x @rolypoly , where r u getting the 2x from dude?

Answer 8

x - x^5 + x^9 + ...
-----------------
1 + x^4 | x
(x + x^5)
---------
-x^5
(-x^5 -x^9)
-----------
x^9
(-x^9 +x^13)
------------
-x^13
etc ...

\[\int(x - x^5 + x^9 + ...)~dx=\frac12x^2 - \frac16x^6 + \frac1{10}x^{10} + ...\]
\[\sum_{0}\frac{(-1)^{n}}{2+4n}x^{2+4n}\]

Answer 9

thanks Jack1. Thing is we were asked to solve for this equation using "u" substitution, and I couldn't do that.
You were asking why 2x right, well if you make x^2= tan$, then when you differentiate both sides, you'll have 2x=sec^2$. Not to sure if I was clear when explaining, maybe someone else has a better explanation.

Answer 10

no, thats clear enough, cheers guys

Answer 11

sweet. :)