# Substitution integral

1. Mar 31, 2004

### 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?

Last edited: Mar 31, 2004
2. Mar 31, 2004

### 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.

3. Mar 31, 2004

### suyver

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

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

4. Mar 31, 2004

### Hurkyl

Staff Emeritus
You could factor the denominator!

5. Apr 1, 2004

### KSCphysics

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

6. Apr 1, 2004

### matt grime

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

7. Apr 1, 2004

### KSCphysics

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

8. Apr 1, 2004

### ahrkron

Staff Emeritus
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.

9. Apr 2, 2004

### 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?