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

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Theorem

$\ds \int \frac {\sqrt {a^2 - x^2} } {x^3} \rd x = \frac {-\sqrt {a^2 - x^2} } {2 x^2} + \frac 1 {2 a} \map \ln {\frac {a + \sqrt {a^2 - x^2} } {\size x} } + C$


Proof

Let:

\(\ds z\) \(=\) \(\ds x^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 x\) Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \int \frac {\sqrt {a^2 - x^2} } {x^3} \rd x\) \(=\) \(\ds \int \frac {\sqrt {a^2 - z} \rd z} {2 z^{3/2} \sqrt z}\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 2 \int \frac {\sqrt {a^2 - z} \rd z} {z^2}\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\frac {-\sqrt {a^2 - z} } z - \frac 1 2 \int \frac {\d z} {z \sqrt {a^2 - z} } } + C\) Primitive of $\dfrac {\sqrt {a x + b} } {x^m}$
\(\ds \) \(=\) \(\ds \frac {-\sqrt {a^2 - x^2} } {2 x^2} - \frac 1 4 \int \frac {2 x \rd x} {x^2 \sqrt {a^2 - x^2} } + C\) substituting for $z$
\(\ds \) \(=\) \(\ds \frac {-\sqrt {a^2 - x^2} } {2 x^2} - \frac 1 2 \int \frac {\d x} {x \sqrt {a^2 - x^2} } + C\) simplifying
\(\ds \) \(=\) \(\ds \frac {-\sqrt {a^2 - x^2} } {2 x^2} - \frac 1 2 \paren {-\frac 1 a \map \ln {\frac {a + \sqrt {a^2 - x^2} } {\size x} } } + C\) Primitive of $\dfrac 1 {x \sqrt {a^2 - x^2} }$
\(\ds \) \(=\) \(\ds \frac {-\sqrt {a^2 - x^2} } {2 x^2} + \frac 1 {2 a} \map \ln {\frac {a + \sqrt {a^2 - x^2} } {\size x} } + C\) simplifying

$\blacksquare$


Also see


Sources