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

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Theorem

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


Proof

Let:

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

$\blacksquare$


Also see


Sources