Calculation of $\int \frac{1}{x^4+1}$

Question

I'm trying to solve this integral in terms of elementary functions but my result does not seem to match up with Sage's or WolframAlpha's.

$$\int \frac{1}{x^4+1} = \frac{x}{1+x^4}+\int \frac{4x^4}{(1+x^4)^2}$$

Where:

$$4\int \frac{x^4}{(1+x^4)^2} = 4\int \frac{u-1}{u^2} = 4\log u + \frac{4}{u}$$

So the original integral is

$$\int \frac{1}{x^4+1} = \frac{x}{1+x^4} + 4\log (1+x^4) + \frac{4}{1+x^4}$$

Where did I go wrong here? Thanks.

You applied a naive change of variable. If we set $u=x^4$, then $du$ is affected, in particular $du = 4x^3 dx$. — Jack D'Aurizio, Oct 01 '16 at 16:31
The trick is just to exploit the Sophie Germain identity $$z^4+4 = (z^2-2z+2)(z^2+2z+2)$$ and partial fraction decomposition. — Jack D'Aurizio, Oct 01 '16 at 16:32
This has been asked so many times. Do check before posting. http://math.stackexchange.com/questions/333611/evaluating-int-frac1x41-dx — StubbornAtom, Oct 01 '16 at 17:59

score 1 · Accepted Answer · answered Oct 01 '16 at 16:32

1

Hint $$x^4+1=(x^2+1)^2-2x^2=(x^2+1-\sqrt{2}x)(x^2+1+\sqrt{2}x)$$

answered Oct 01 '16 at 16:32

E.H.E

23,280

Aakash Kumar · Answer 2 · 2016-10-01T16:50:32.370

1

$$I=\frac12\int\frac {\frac{2}{x^2}}{x^2+\frac {1}{x^2}} $$ $$I=\frac12\int\frac{1+\frac {1}{x^2}}{(x-\frac {1}{x})^2+2}{dx}-\frac12\int\frac{1-\frac {1}{x^2}}{(x+\frac {1}{x})^2-2}{dx}$$

edited Oct 01 '16 at 16:50

answered Oct 01 '16 at 16:47

Aakash Kumar

3,480

Calculation of $\int \frac{1}{x^4+1}$

2 Answers2