Primitive of Root of x squared plus a squared cubed over x cubed
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Theorem
- $\ds \int \frac {\paren {\sqrt {x^2 + a^2} }^3} {x^3} \rd x = \frac {-\paren {\sqrt {x^2 + a^2} }^3} {2 x^2} + \frac {3 \sqrt {x^2 + a^2} } 2 - \frac {3 a} 2 \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C$
Proof
Let:
\(\ds z\) | \(=\) | \(\ds x^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d z} {\d x}\) | \(=\) | \(\ds 2 x\) | Power Rule for Derivatives | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \frac {\paren {\sqrt {x^2 + a^2} }^3} {x^3} \rd x\) | \(=\) | \(\ds \int \frac {\paren {\sqrt {z + a^2} }^3} {2 z^2} \rd z\) | Integration by Substitution | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \paren {\frac {-\paren {\sqrt {z + a^2} }^3} z + \frac 3 2 \int \frac {\sqrt {z + a^2} } z \rd z}\) | Primitive of $\dfrac {\paren {a x + b}^m} {\paren {p x + q}^n}$: Formulation 3 | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-\paren {\sqrt {x^2 + a^2} }^3} {2 x^2} + \frac 3 2 \int \frac {\sqrt {x^2 + a^2} } x \rd x\) | substituting for $z$ and simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-\paren {\sqrt {x^2 + a^2} }^3} {2 x^2} + \frac 3 2 \paren {\sqrt {x^2 + a^2} - a \map \ln {\frac {a + \sqrt {x^2 + a^2} } a} } + C\) | Primitive of $\dfrac {\sqrt {x^2 + a^2} } x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-\paren {\sqrt {x^2 + a^2} }^3} {2 x^2} + \frac {3 \sqrt {x^2 + a^2} } 2 - \frac {3 a} 2 \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C\) | simplifying |
$\blacksquare$
Also see
- Primitive of $\dfrac {\paren {\sqrt {x^2 - a^2} }^3} {x^3}$
- Primitive of $\dfrac {\paren {\sqrt {a^2 - x^2} }^3} {x^3}$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {x^2 + a^2}$: $14.209$