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?
 A: Alternate forms exist. According to Wolfram Alpha:


The first one and the second one are the same because:
$$
\tanh^{-1}(x):=\frac12\log\left(\frac{1+x}{1-x}\right)
$$
A: 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.
A: $\boxed {F(x) =\dfrac{1}{4} \ln|x-1| - \dfrac{1}{4} \ln|x+1| + \dfrac{1}{2} \arctan{x} + C}$ 
F is defined over $\mathbb{R}\backslash \{-1,1\}$ 
Regarding your result, Your function is defined only over $]-1,1[$, why?
Now for the message from Bernard, thank you, I realised I must used the u-substitution. So :
$\displaystyle {F(x)=\int \dfrac{dx}{1-x^4} \implies F(u)=\int \dfrac{du}{2\sqrt{u}(1-u^2)} ,\quad (\text{with} \; u=x^2)}$
This form is not interesting.
Thanks every one
