Primitive of Reciprocal of Root of x squared minus a squared cubed
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Theorem
- $\ds \int \frac {\d x} {\paren {\sqrt {x^2 - a^2} }^3} = \frac {-x} {a^2 \sqrt {x^2 - a^2} } + C$
for $\size x > a$.
Proof
Let:
\(\ds z\) | \(=\) | \(\ds x^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d z} {\d x}\) | \(=\) | \(\ds 2 x\) | Power Rule for Derivatives | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \frac {\d x} {\paren {\sqrt {x^2 - a^2} }^3}\) | \(=\) | \(\ds \int \frac {\d z} {2 \sqrt z \paren {\sqrt {z - a^2} }^3}\) | 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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \paren {\frac {2 \sqrt z} {\paren {-a^2} \sqrt {z - a^2} } } + C\) | Primitive of $\dfrac 1 {\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} } }$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \paren {\frac {2 x} {\paren {-a^2} \sqrt {x^2 - a^2} } } + C\) | substituting for $z$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-x} {a^2 \sqrt {x^2 - a^2} } + C\) | simplifying |
$\blacksquare$
Also see
- Primitive of $\dfrac 1 {\paren {\sqrt {x^2 + a^2} }^3}$
- Primitive of $\dfrac 1 {\paren {\sqrt {a^2 - x^2} }^3}$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {x^2 - a^2}$: $14.223$