Question

need help solving the integral of x/(1+x^4)dx.

Answer 1

\[\int\limits_{}^{}x/(1+x^4)dx\]

Answer 2

x = tan^2 ?

Answer 3

replace
\[u=x^2\] get inverse tangent in one step

Answer 4

thats the thought ... :)

Answer 5

but then x = sqrt(u)

Answer 6

that is if you remember that
\[\frac{d}{dx}\tan^{-1}(x)=\frac{1}{1+x^2}\]

Answer 7

first use
u = x^2
du = 2x dx

then use trig substitution
u = tan(theta)
du = sec^2

Answer 8

\[u=x^2\]
\[du=2xdx\]
\[\frac{1}{2}\int\frac{1}{1+u^2}du=\frac{1}{2}\tan^{-1}(u)\]]

Answer 9

i sometimes forget to convert the dx:du part :)

Answer 10

you can use a trig sub if you like, but if you already know that the derivative of arctangent is
\[\frac{1}{1+x^2}\] then it is not necessary

Answer 11

oh so "final answer" is
\[\frac{1}{2}\tan^{-1}(x^2)\] and don't forget the stupid "plus cee"

Answer 12

Thanks, everyone, for the help :) I had started it out, but it seemed too easy, so thought I'd double check. It's not a complicated prob at all...