<s>[INST] 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?
 [/INST] $\boxed {F(x) =\dfrac{1}{4} \ln|x-1| - \dfrac{1}{4} \ln|x+1| + \dfrac{1}{2} \arctan{x} + C}$ 
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
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