Primitive of Minus One plus x Squared over One plus Fourth Power of x

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Theorem

$\ds \int \frac {x^2 - 1} {x^4 + 1} \rd x = \frac 1 {2 \sqrt 2} \ln \size {\frac {x^2 - \sqrt 2 x + 1} {x^2 + \sqrt 2 x + 1} } + C$


Proof 1

We have:

\(\ds \int \frac {x^2 - 1} {x^4 + 1} \rd x\) \(=\) \(\ds \int \frac {1 - \frac 1 {x^2} } {x^2 + \frac 1 {x^2} } \rd x\)
\(\ds \) \(=\) \(\ds \int \frac {1 - \frac 1 {x^2} } {\paren {x + \frac 1 x}^2 - 2} \rd x\) Completing the Square

Note that, by Derivative of Power:

$\dfrac \d {\d x} \paren {x + \dfrac 1 x} = 1 - \dfrac 1 {x^2}$

So, we have:

\(\ds \int \frac {1 - \frac 1 {x^2} } {\paren {x + \frac 1 x}^2 - 2} \rd x\) \(=\) \(\ds \int \frac 1 {u^2 - 2} \rd u\) substituting $u = x + \dfrac 1 x$
\(\ds \) \(=\) \(\ds \frac 1 {2 \sqrt 2} \ln \size {\frac {u - \sqrt 2} {u + \sqrt 2} } + C\) Primitive of $\dfrac 1 {x^2 - a^2}$: Logarithm Form
\(\ds \) \(=\) \(\ds \frac 1 {2 \sqrt 2} \ln \size {\frac {x + \frac 1 x - \sqrt 2} {x + \frac 1 x + \sqrt 2} } + C\)
\(\ds \) \(=\) \(\ds \frac 1 {2 \sqrt 2} \ln \size {\frac {x^2 - \sqrt 2 x + 1} {x^2 + \sqrt 2 x + 1} } + C\)

$\blacksquare$


Proof 2

\(\ds \int \frac {x^2 - 1} {x^4 + 1} \rd x\) \(=\) \(\ds \int \frac {x^2} {x^4 + 1} \rd x - \int \frac 1 {x^4 + 1} \rd x\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 {4 \sqrt 2} \map \ln {\frac {x^2 - x \sqrt 2 + 1} {x^2 + x \sqrt 2 + 1} } - \frac 1 {2 \sqrt 2} \paren {\map \arctan {1 - x \sqrt 2} - \map \arctan {1 + x \sqrt 2} }\) Primitive of $\dfrac {x^2} {x^4 + a^4}$, setting $a = 1$
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac 1 {2 \sqrt 2} \paren {\map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } + \frac 1 2 \ln \size {\frac {x^2 + \sqrt 2 x + 1} {x^2 - \sqrt 2 x + 1} } }\) Primitive of $\dfrac 1 {1 + x^4}$