Primitive of Reciprocal of x cubed by Root of x squared plus a squared
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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
- Primitive of Reciprocal of $x^3 \sqrt {x^2 - a^2}$
- Primitive of Reciprocal of $x^3 \sqrt {a^2 - x^2}$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {x^2 + a^2}$: $14.188$