Primitive of x over a squared minus x squared

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Theorem

$\displaystyle \int \frac {x \rd x} {a^2 - x^2} = -\frac 1 2 \, \map \ln {a^2 - x^2} + C$

for $x^2 < a^2$.


Proof

Let:

\(\displaystyle z\) \(=\) \(\displaystyle a^2 - x^2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\d z} {\d x}\) \(=\) \(\displaystyle -2 x\) Power Rule for Derivatives
\(\displaystyle \leadsto \ \ \) \(\displaystyle \int \frac {\d x} {x \paren {a^2 - x^2} }\) \(=\) \(\displaystyle \int \frac {\d z} {-2 z}\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle -\frac 1 2 \int \frac {\d z} z\) Primitive of Constant Multiple of Function
\(\displaystyle \) \(=\) \(\displaystyle -\frac 1 2 \ln z + C\) Primitive of Reciprocal: Corollary as $z > 0$
\(\displaystyle \) \(=\) \(\displaystyle -\frac 1 2 \, \map \ln {a^2 - x^2} + C\) substituting for $z$

$\blacksquare$


Also see


Sources