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

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Theorem

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

for $\size x > a$.


Proof 1

Let:

\(\ds x\) \(=\) \(\ds a \cosh \theta\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d x} {\d \theta}\) \(=\) \(\ds a \sinh \theta\) Derivative of Hyperbolic Cosine


Then:

\(\ds x\) \(=\) \(\ds a \cosh \theta\)
\(\ds \leadsto \ \ \) \(\ds \sqrt {x^2 - a^2}\) \(=\) \(\ds \sqrt {a^2 \paren {\cosh^2 \theta - 1} }\)
\(\ds \) \(=\) \(\ds \sqrt {a^2 \sinh^2 \theta}\) Difference of Squares of Hyperbolic Cosine and Sine
\(\ds \) \(=\) \(\ds a \sinh \theta\)


Hence:

\(\ds \int \frac {\d x} {x^2 \sqrt {x^2 - a^2} }\) \(=\) \(\ds \int \frac {a \sinh \theta \rd \theta} {a^2 \cosh^2 \theta \cdot a \sinh \theta}\) Integration by Substitution
\(\ds \) \(=\) \(\ds \dfrac 1 {a^2} \int \frac {\rd \theta} {\cosh^2 \theta}\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 1 {a^2} \tanh \theta + C\) Primitive of $\dfrac 1 {\cosh^2 \theta}$
\(\ds \) \(=\) \(\ds \dfrac 1 {a^2} \dfrac {a \sinh \theta} {a \cosh \theta} + C\) Definition 2 of Hyperbolic Tangent
\(\ds \) \(=\) \(\ds \dfrac 1 {a^2} \dfrac {\sqrt {x^2 - a^2} } x + C\) substituting for $a \sinh \theta$ and $a \cosh \theta$

$\blacksquare$


Proof 2

Let:

\(\ds z\) \(=\) \(\ds x^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 x\) Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {x^2 \sqrt {x^2 - a^2} }\) \(=\) \(\ds \int \frac {\d z} {2 z \sqrt z \sqrt {z - a^2} }\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 2 \int \frac {\d z} {z^{3/2} \sqrt {z - a^2} }\)


Using Primitive of $\dfrac 1 {u^m \sqrt {a u + b} }$:

$\ds \int \frac {\d u} {u^m \sqrt {a u + b} } = -\frac {\sqrt {a u + b} } {\paren {m - 1} b u^{m - 1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d u} {u^{m - 1} \sqrt {a u + b} }$


Setting:

\(\ds u\) \(:=\) \(\ds z\)
\(\ds m\) \(:=\) \(\ds \frac 3 2\)
\(\ds a\) \(:=\) \(\ds 1\)
\(\ds b\) \(:=\) \(\ds -a^2\)


\(\ds \frac 1 2 \int \frac {\d z} {z^{3/2} \sqrt {z - a^2} }\) \(=\) \(\ds \frac {-\sqrt {z - a^2} } {2 \paren {\paren {\frac 3 2} - 1} \paren {-a^2} z^{\paren {3/2} - 1} } - \frac {2 \paren {\frac 3 2} - 3} {2 \paren {2 \paren {\frac 3 2} - 2} \paren {-a^2} } \int \frac {\d z} {z^{\paren {\frac 3 2} - 1} \sqrt{z - a^2} } + C\) Primitive of $ \dfrac 1 {x^m \sqrt {a x + b} }$
\(\ds \) \(=\) \(\ds \frac {-\sqrt {z - a^2} } {-a^2 z^{1/2} } - 0 + C\) simplifying
\(\ds \) \(=\) \(\ds \frac {\sqrt {x^2 - a^2} } {a^2 x} + C\) substituting $x$ back for $z$

$\blacksquare$


Also see


Sources

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