Primitive of Reciprocal of x fourth plus a fourth/Lemma 2
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Lemma for Primitive of Reciprocal of $x^4 + a^4$
- $\ds \int \frac {\d x} {x^2 - a x \sqrt 2 + a^2} = \frac {-\sqrt 2} a \map \arctan {1 - \frac {x \sqrt 2} a}$
Proof
The discriminant of $x^2 - a x \sqrt 2 + a^2$ is:
\(\ds \map {\operatorname {Disc} } {x^2 - a x \sqrt 2 + a^2}\) | \(=\) | \(\ds \paren {-a \sqrt 2}^2 - 4 \times 1 \times a^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 a^2 - 4 a^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds - 2 a^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds < 0\) |
Thus:
\(\ds \int \frac {\d x} {x^2 - a x \sqrt 2 + a^2}\) | \(=\) | \(\ds \frac 2 {\sqrt {4 a^2 - \paren {-a \sqrt 2}^2} } \map \arctan {\frac {2 x - a \sqrt 2} {\sqrt {4 a^2 - \paren {-a \sqrt 2}^2} } }\) | Primitive of $\dfrac 1 {a x^2 + b x + c}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {a \sqrt 2} \map \arctan {\frac {2 x - a \sqrt 2} {a \sqrt 2} }\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\sqrt 2} a \map \arctan {\frac {x \sqrt 2} a - 1}\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-\sqrt 2} a \map \arctan {1 - \frac {x \sqrt 2} a}\) | Arctangent is Odd Function |
$\blacksquare$