Primitive of Reciprocal of x fourth plus a fourth/Lemma 1
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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 | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\sqrt 2} a \map \arctan {1 + \frac {x \sqrt 2} a}\) | simplifying |
$\blacksquare$