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

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Proof

Let:

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


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

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


Setting:

\(\ds x\) \(:=\) \(\ds z\)
\(\ds m\) \(:=\) \(\ds 2\)
\(\ds a\) \(:=\) \(\ds 1\)
\(\ds b\) \(:=\) \(\ds a^2\)


\(\ds \frac 1 2 \int \frac {\d z} {z^2 \sqrt {z + a^2} }\) \(=\) \(\ds \frac {-\sqrt {z + a^2} } {2 \paren {2 - 1} a^2 z^{2 - 1} } - \frac {2 \paren 2 - 3} {2 \paren {2 \paren 2 - 2} a^2} \int \frac {\d z} {z^{2 - 1} \sqrt {z + a^2} } + C\) Primitive of $ \dfrac 1 {x^m \sqrt {a x + b} }$
\(\ds \) \(=\) \(\ds \frac {-\sqrt {z + a^2} } {2 a^2 z} - \frac 1 {4 a^2} \int \frac {\d z} {z \sqrt {z + a^2} } + C\) simplifying
\(\ds \) \(=\) \(\ds \frac {-\sqrt {x^2 + a^2} } {2 a^2 x^2} - \frac 1 {4 a^2} \paren {\int \frac {2 x \rd x} {x^2 \sqrt {x^2 + a^2} } } + C\) substituting for $z$
\(\ds \) \(=\) \(\ds \frac {-\sqrt {x^2 + a^2} } {2 a^2 x^2} - \frac 1 {2 a^2} \paren {\int \frac {\d x} {x \sqrt {x^2 + a^2} } } + C\) simplification
\(\ds \) \(=\) \(\ds \frac {-\sqrt {x^2 + a^2} } {2 a^2 x^2} + \frac 1 {2 a^3} \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C\) Primitive of $\dfrac 1 {x \sqrt {x^2 + a^2} }$

$\blacksquare$


Also see


Sources