Primitive of Reciprocal of Root of x squared minus a squared cubed

From ProofWiki
Jump to navigation Jump to search

Theorem

$\ds \int \frac {\d x} {\paren {\sqrt {x^2 - a^2} }^3} = \frac {-x} {a^2 \sqrt {x^2 - a^2} } + C$

for $\size x > a$.


Proof

Let:

\(\ds z\) \(=\) \(\ds x^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 x\) Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {\paren {\sqrt {x^2 - a^2} }^3}\) \(=\) \(\ds \int \frac {\d z} {2 \sqrt z \paren {\sqrt {z - a^2} }^3}\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 2 \int \frac {\d z} {\paren {z - a^2} \sqrt z \sqrt {z - a^2} }\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\frac {2 \sqrt z} {\paren {-a^2} \sqrt {z - a^2} } } + C\) Primitive of $\dfrac 1 {\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} } }$
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\frac {2 x} {\paren {-a^2} \sqrt {x^2 - a^2} } } + C\) substituting for $z$
\(\ds \) \(=\) \(\ds \frac {-x} {a^2 \sqrt {x^2 - a^2} } + C\) simplifying

$\blacksquare$


Also see


Sources