Integral

[Jupiter]

\int\frac{dx}{x^4+1}
What really frustrates me is that I've seen this integral before. I believe it involved some whacky subsitution like x=e^u, but no substitution seems to work. Partial fractions just make a mess. Trig subs seem tempting but that 4 screws everything up. Ideas?

[matt grime]

If you do that whacky substitution, how about then multiplying through to get e^(2u)+e^(-2u) and getting a hyperbolic trig function? Not saying this works.

[suyver]

\int\frac{dx}{x^4+1}=\frac{1}{3\sqrt{2}}
\left(2\arctan(1+\sqrt{2}x)-2\arctan(1-\sqrt{2}x)+\right.

\left.\log(\sqrt{2}x+x^2+1)-\log(\sqrt{2}x-x^2-1)\right)

[Hurkyl]

You could factor the denominator!

[KSCphysics]

isn't the integral of 1/x^4+1 just the ln|x^4+1|??? or something like that?

[matt grime]

No, differentiate log(x^4+1) and you'll see why.

[KSCphysics]

noooo. i mean, the integral of 1/u is the ln|u|+c and in this case, 1/u = (x^4+1)

[ahrkron]

In order to do that, you also need to have a full du on your integrand, which would need an x^3 term that is not there.

[matt grime]

Also not that if you're correct that the integral of 1/x^2 is log(x^2) and, obviously -1/x as well, so up to a constant 2logx = -1/x?