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