Primitive of x over a squared minus x squared squared
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Theorem
- $\ds \int \frac {x \rd x} {\paren {a^2 - x^2}^2} = \frac 1 {2 \paren {a^2 - x^2} } + C$
for $x^2 < a^2$.
Proof
Let:
\(\ds z\) | \(=\) | \(\ds a^2 - x^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d z} {\d x}\) | \(=\) | \(\ds -2 x\) | Derivative of Power | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \frac {x \rd x} {\paren {a^2 - x^2}^2}\) | \(=\) | \(\ds \int \frac {\d z} {-2 z^2}\) | Integration by Substitution | ||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 2 \int \frac {\d z} {z^2}\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 2 \paren {\frac {-1} z} + C\) | Primitive of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 \paren {a^2 - x^2} } + C\) | substituting for $z$ and simplifying |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.171$