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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?

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

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  • 1
    $\begingroup$ Yes, You are right !! $\endgroup$ – alkabary Jun 2 '15 at 21:34
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    $\begingroup$ @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 $d(x^2)=2\mathbf{x}dx$ ? $\endgroup$ – Alexey Burdin Jun 2 '15 at 21:37
  • $\begingroup$ I'm with @AlexeyBurdin, your $u$-substitution is very fishy. $\endgroup$ – Alex Schiff Jun 2 '15 at 21:39
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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.

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  • $\begingroup$ oh, my God. You are right. I made a mistake. $\endgroup$ – apollonian Jun 3 '15 at 17:16

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