Primitive of Root of x squared plus a squared over x squared/Logarithm Form
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Theorem
- $\ds \int \frac {\sqrt {x^2 + a^2} } {x^2} \rd x = \frac {-\sqrt {x^2 + a^2} } x + \map \ln {x + \sqrt {x^2 + a^2} } + C$
Proof
Let:
\(\ds z\) | \(=\) | \(\ds x^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d z} {\d x}\) | \(=\) | \(\ds 2 x\) | Power Rule for Derivatives | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \frac {\sqrt {x^2 + a^2} } {x^2} \rd x\) | \(=\) | \(\ds \int \frac {\sqrt {z + a^2} \rd z} {2 z \sqrt z}\) | Integration by Substitution | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \int \frac {\sqrt {z + a^2} \rd z} {z^{3/2} }\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \paren {\frac {-\sqrt {z + a^2} } {\frac 1 2 \sqrt z} + \frac 1 2 \int \frac {\d z} {\sqrt z \sqrt {z + a^2} } } + C\) | Primitive of $\dfrac {\sqrt {a x + b} } {x^m}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-\sqrt {x^2 + a^2} } x + \frac 1 2 \int \frac {2 x \rd x} {x \sqrt {x^2 + a^2} } + C\) | substituting for $z$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-\sqrt {x^2 + a^2} } x + \int \frac {\d x} {\sqrt {x^2 + a^2} } + C\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-\sqrt {x^2 + a^2} } x + \map \ln {x + \sqrt {x^2 + a^2} } + C\) | Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$ |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {x^2 + a^2}$: $14.194$