Primitive of Reciprocal of x squared by Root of x squared plus a squared cubed
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Theorem
- $\ds \int \frac {\d x} {x^2 \paren {\sqrt {x^2 + a^2} }^3} = \frac {-\sqrt {x^2 + a^2} } {a^4 x} - \frac x {a^4 \sqrt {x^2 + a^2} } + C$
Proof
\(\ds \int \frac {\d x} {x^2 \paren {\sqrt {x^2 + a^2} }^3}\) | \(=\) | \(\ds \int \frac {a^2 \rd x} {a^2 x^2 \paren {\sqrt {x^2 + a^2} }^3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {\paren {x^2 + a^2 - x^2} \rd x} {a^2 x^2 \paren {\sqrt {x^2 + a^2} }^3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {a^2} \int \frac {\paren {x^2 + a^2} \rd x} {a^2 x^2 \paren {\sqrt {x^2 + a^2} }^3} - \frac 1 {a^2} \int \frac {x^2 \rd x} {x^2 \paren {\sqrt {x^2 + a^2} }^3}\) | Linear Combination of Primitives | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {a^2} \int \frac {\d x} {x^2 \sqrt {x^2 + a^2} } - \frac 1 {a^2} \int \frac {\d x} {\paren {\sqrt {x^2 + a^2} }^3}\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {a^2} \frac {-\sqrt {x^2 + a^2} } {a^2 x} - \frac 1 {a^2} \int \frac {\d x} {\paren {\sqrt {x^2 + a^2} }^3} + C\) | Primitive of $\dfrac 1 {x^2 \sqrt {x^2 + a^2} }$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-\sqrt {x^2 + a^2} } {a^4 x} - \frac 1 {a^2} \frac x {a^2 \sqrt {x^2 + a^2} } + C\) | Primitive of $\dfrac 1 {\paren {\sqrt {x^2 + a^2} }^3}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-\sqrt {x^2 + a^2} } {a^4 x} - \frac x {a^4 \sqrt {x^2 + a^2} } + C\) | simplification |
$\blacksquare$
Also see
- Primitive of $\dfrac 1 {x^2 \paren {\sqrt {x^2 - a^2} }^3}$
- Primitive of $\dfrac 1 {x^2 \paren {\sqrt {a^2 - x^2} }^3}$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {x^2 + a^2}$: $14.201$