What's the relation between different antiderivatives? If a function $f(x)$ has different forms of antiderivatives:
$\frac { d }{ dx } { F }_{ 1 }(x)=f(x)$
$\frac { d }{ dx } { F }_{ 2 }(x)=f(x)$
What's the relationship between $F_1$ and $F_2$, is that ${F}_{1}(x)-{F}_{2}(x)=constant$ correct?
For example, question find $\int { \frac { dx }{ { x }^{ 4 }-1 } = } $ ?
Method 1: $\int { \frac { dx }{ { x }^{ 4 }-1 } =\int { \frac { dx }{ \left( { x }^{ 2 }-1 \right) \left( { x }^{ 2 }+1 \right)  } =\frac { 1 }{ 2 } \int { \frac { dx }{ { x }^{ 2 }-1 } -\frac { 1 }{ 2 } \int { \frac { dx }{ { x }^{ 2 }+1 }  } =\frac { 1 }{ 4 } ln\left| \frac { x-1 }{ x+1 }  \right| -\frac { 1 }{ 2 } arctan(x) } +c }  } $
Method 2:$\int { \frac { dx }{ { x }^{ 4 }-1 } =\frac { 1 }{ 2 } \int { \frac { d{ x }^{ 2 } }{ { \left( { x }^{ 2 } \right)  }^{ 2 }-1 } =\frac { 1 }{ 2 } ln\left| \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 }  \right| +c }  } $
Ok, now the question is: what's the relation between $ln\left| \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 }  \right| $ and $\frac { 1 }{ 2 } ln\left| \frac { { x }-1 }{ { x }+1 }  \right| -arctan(x)$ ?
Does the equation below is correct and how to prove it?
$ln\left| \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 }  \right| =\frac { 1 }{ 2 } ln\left| \frac { { x }-1 }{ { x }+1 }  \right| -arctan(x) +constant$ 

Your method 1 is correct.
Your method 2 is not:
$$\int\frac{dx}{x^4-1} \neq \frac12\int\frac{d(x^2)}{(x^2)^2-1}$$
If you're going to substitute for $x^2$, $d(x^2) = 2x dx$, so the equality would be
$$\int\frac{dx}{x^4-1} = \frac12\int\frac{du}{\sqrt{u}(u^2-1)}$$
which gets you nowhere. 
But had both methods been correct, yes, the two anti-derivatives would differ by a constant. 
