# 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?

$ln\left| \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } \right| =\frac { 1 }{ 2 } ln\left| \frac { { x }-1 }{ { x }+1 } \right| -arctan(x) +constant$

• Yes, You are right !! – alkabary Jun 2 '15 at 21:34
• @apollonian: How did you $\int \frac{ dx }{ { x }^{ 4 }-1 } = \frac{ 1 }{ 2 } \int \frac{ dx^2 }{ \left( x^2 \right)^2-1}$ ? Isn't $d(x^2)=2\mathbf{x}dx$ ? – Alexey Burdin Jun 2 '15 at 21:37
• I'm with @AlexeyBurdin, your $u$-substitution is very fishy. – Alex Schiff Jun 2 '15 at 21:39

$$\int\frac{dx}{x^4-1} \neq \frac12\int\frac{d(x^2)}{(x^2)^2-1}$$
If you're going to substitute for $x^2$, $d(x^2) = 2x dx$, so the equality would be
$$\int\frac{dx}{x^4-1} = \frac12\int\frac{du}{\sqrt{u}(u^2-1)}$$