Primitive of Root of x squared plus a squared cubed over x
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Theorem
- $\ds \int \frac {\paren {\sqrt {x^2 + a^2} }^3} x \rd x = \frac {\paren {\sqrt {x^2 + a^2} }^3} 3 + a^2 \sqrt {x^2 + a^2} - a^3 \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C$
Proof
Let:
\(\ds z\) | \(=\) | \(\ds x^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d z} {\d x}\) | \(=\) | \(\ds 2 x\) | Power Rule for Derivatives | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \frac {\paren {\sqrt {x^2 + a^2} }^3} x \rd x\) | \(=\) | \(\ds \int \frac {\paren {\sqrt {z + a^2} }^3} {2 z} \rd z\) | Integration by Substitution | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {\sqrt {z + a^2} }^3} 3 + \frac {a^2} 2 \int \frac {\sqrt {z + a^2} } z \rd z\) | Primitive of $\dfrac {\paren {\sqrt {a x + b} }^m} x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {\sqrt {z + a^2} }^3} 3 + \frac {a^2} 2 \paren {2 \sqrt {z + a^2} + a^2 \int \frac {\d z} {z \sqrt {z + a^2} } }\) | Primitive of $\dfrac {\sqrt {a x + b} } x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {\sqrt {x^2 + a^2} }^3} 3 + a^2 \sqrt {x^2 + a^2} + a^4 \int \frac {\d x} {x \sqrt {x^2 + a^2} }\) | substituting for $z$ and simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {\sqrt {x^2 + a^2} }^3} 3 + a^2 \sqrt {x^2 + a^2} - a^3 \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C\) | Primitive of $\dfrac 1 {x \sqrt {x^2 + a^2} }$ |
$\blacksquare$
Also see
- Primitive of $\dfrac {\paren {\sqrt {x^2 - a^2} }^3} x$
- Primitive of $\dfrac {\paren {\sqrt {a^2 - x^2} }^3} x$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {x^2 + a^2}$: $14.207$