Primitive of x over Root of x squared plus a squared cubed
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Theorem
- $\ds \int \frac {x \rd x} {\paren {\sqrt {x^2 + a^2} }^3} = \frac {-1} {\sqrt {x^2 + a^2} } + C$
Proof
Let:
\(\ds z\) | \(=\) | \(\ds x^2 + a^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d z} {\d x}\) | \(=\) | \(\ds 2 x\) | Power Rule for Derivatives | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \frac {x \rd x} {\paren {\sqrt {x^2 + a^2} }^3}\) | \(=\) | \(\ds \int \frac {x \rd z} {2 x z^{3/2} }\) | Integration by Substitution | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \int z^{-3/2} \rd z\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \paren {\frac {-z^{1/2} } {\frac 1 2} } + C\) | Primitive of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-1} {\sqrt z} + C\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-1} {\sqrt {x^2 + a^2} } + C\) | substituting for $z$ |
$\blacksquare$
Also see
- Primitive of $\dfrac x {\paren {\sqrt {x^2 - a^2} }^3}$
- Primitive of $\dfrac x {\paren {\sqrt {a^2 - x^2} }^3}$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {x^2 + a^2}$: $14.197$