Primitive of Minus One plus x Squared over One plus Fourth Power of x/Proof 1

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

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

\(\ds \)

\(=\)

\(\ds \frac 1 {2 \sqrt 2} \ln \size {\frac {x + \frac 1 x - \sqrt 2} {x + \frac 1 x + \sqrt 2} } + C\)

\(\ds \)

\(=\)

\(\ds \frac 1 {2 \sqrt 2} \ln \size {\frac {x^2 - \sqrt 2 x + 1} {x^2 + \sqrt 2 x + 1} } + C\)

$\blacksquare$