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

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Theorem

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


Proof 1

\(\ds x\) \(=\) \(\ds a \tan \theta\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d x} {\d \theta}\) \(=\) \(\ds a \sec^2 \theta\) Derivative of Tangent Function
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {\paren {\sqrt {x^2 + a^2} }^3}\) \(=\) \(\ds \int \frac {a \sec^2 \theta \rd \theta} {\sqrt {a^2 \tan^2 \theta + a^2}^3}\) Integration by Substitution
\(\ds \) \(=\) \(\ds \int \frac {a \sec^2 \theta \rd \theta} {a^3 \sec^3 \theta}\) Difference of Squares of Secant and Tangent
\(\ds \) \(=\) \(\ds \dfrac 1 {a^2} \int \cos \theta \rd \theta\) Definition of Real Secant Function and simplification
\(\ds \) \(=\) \(\ds \dfrac 1 {a^2} \sin \theta + C\) Primitive of $\cos x$
\(\ds \) \(=\) \(\ds \dfrac 1 {a^2} \dfrac {a \sin \theta} {\cos \theta} \dfrac {\cos \theta} a + C\)
\(\ds \) \(=\) \(\ds \dfrac 1 {a^2} \dfrac {a \tan \theta} {a \sec \theta} + C\) Tangent is Sine divided by Cosine, Definition of Real Secant Function
\(\ds \) \(=\) \(\ds \dfrac 1 {a^2} \dfrac {a \tan \theta} {\sqrt {a^2 \sec^2 \theta} } + C\)
\(\ds \) \(=\) \(\ds \dfrac 1 {a^2} \dfrac {a \tan \theta} {\sqrt {a^2 \tan^2 \theta + a^2} } + C\) Difference of Squares of Secant and Tangent
\(\ds \) \(=\) \(\ds \dfrac 1 {a^2} \dfrac x {\sqrt {x^2 + a^2} } + C\) substituting $x = a \tan \theta$

$\blacksquare$


Proof 2

\(\ds z\) \(=\) \(\ds x^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 x\) 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
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\frac {2 \sqrt z} {a^2 \sqrt {z + a^2} } } + C\) Primitive of $\dfrac 1 {\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} } }$
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\frac {2 x} {a^2 \sqrt {x^2 + a^2} } } + C\) substituting for $z$
\(\ds \) \(=\) \(\ds \frac x {a^2 \sqrt {x^2 + a^2} } + C\) simplifying

$\blacksquare$


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