Two ways to calculate the primitive of $f(x) =\frac{x^2}{x^4-1}$ but two different results, why? When I search  primitives of $f(x)= \dfrac{x^2}{x^4-1}$, using partial fraction decomposition the solutions are 
$\boxed {F(x) =\dfrac{1}{4} \ln|x-1| + \dfrac{1}{4} \ln|x+1| + \dfrac{1}{2} \arctan{x} + C}$
But if I use an other way, I've got an another result :
$f(x)=\dfrac{x^2}{x^4-1} =\dfrac{x^2-1+1}{(x^2-1)(x^2+1)} =\dfrac{x^2-1}{(x^2-1)(x^2+1)}+\dfrac{1}{(x^2-1)(x^2+1)} = \\\dfrac{1}{x^2+1} +\dfrac{1}{x^4-1}= \dfrac{1}{x^2+1} -\dfrac{1}{1-(x^2)^2}$
Thus $\boxed {F(x)= \arctan{x} -\operatorname{argth}x^2 +C} $
I think that might be wrong, because that doesn't match with the first result if I plot both on geogebra. I noticed as well their domains are different.
My question is why this two different results? Is one of them wrong?

There is nothing strange in having different expressions for an antiderivative: an antiderivative of $1/x$ over $(0,\infty)$ is $\log x$ as well as $\log(3x)$, because antiderivatives are determined (over an interval) up to an additive constant. However, your second computation is wrong.
An antiderivative of $1/(1-x^2)$ is $\operatorname{artanh}x$ (or, with your notation, $\operatorname{argth}x$), but then it's false that $\operatorname{artanh}(x^2)$ is an antiderivative of $1/(1-x^4)$: indeed
$$
\frac{d}{dx}\operatorname{artanh}(x^2)=
\frac{1}{1-(x^2)^2}\cdot 2x
$$
by the chain rule.
The mistake would be the same as saying that, since an antiderivative of $x$ is $x^2/2$, an antiderivative of $x^4=(x^2)^2$ is $(x^2)^2/2$, which is clearly wrong.
