Primitive of Reciprocal of x cubed plus a cubed/Lemma
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Lemma for Primitive of Reciprocal of $x^3 + a^3$
- $\ds \int \frac {\d x} {x^2 - a x + a^2} = \frac 2 {a \sqrt 3} \map \arctan {\frac {2 x - a} {a \sqrt 3} }$
Proof
The discriminant of $x^2 - a x + a^2$ is:
\(\ds \map {\mathrm {Disc} } {x^2 - a x + a^2}\) | \(=\) | \(\ds \paren {-a}^2 - 4 \times 1 \times a^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a^2 - 4 a^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -3 a^2\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds 0\) |
Thus:
\(\ds \int \frac {\d x} {x^2 - a x + a^2}\) | \(=\) | \(\ds \frac 2 {\sqrt {4 a^2 - \paren {-a}^2} } \map \arctan {\frac {2 x - a} {\sqrt {4 a^2 - \paren {-a}^2} } }\) | Primitive of $\dfrac 1 {a x^2 + b x + c}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {a \sqrt 3} \map \arctan {\frac {2 x - a} {a \sqrt 3} }\) | simplifying |
$\blacksquare$