Primitive of Reciprocal of x by Root of Power of x plus Power of a

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Theorem

$\ds \int \frac {\d x} {x \sqrt {x^n + a^n} } = \frac 1 {n \sqrt {a^n} } \ln \size {\frac {\sqrt {x^n + a^n} - \sqrt {a^n} } {\sqrt {x^n + a^n} + \sqrt {a^n} } } + C$


Proof

\(\ds u\) \(=\) \(\ds \sqrt {x^n + a^n}\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds \frac {n x^{n - 1} } {2 \sqrt {x^n + a^n} }\) Derivative of Power, Chain Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {x \sqrt {x^n + a^n} }\) \(=\) \(\ds \int \frac {2 \sqrt {x^n + a^n} \rd u} {n x^{n - 1} x \sqrt {x^n + a^n} }\) Integration by Substitution
\(\ds \) \(=\) \(\ds \int \frac {2 \rd u} {n \paren {u^2 - a^n} }\) completing substitution and simplifying
\(\ds \) \(=\) \(\ds \frac 2 n \int \frac {\d u} {\paren {u^2 - \paren {\sqrt {a^n} } ^2} }\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 2 n \paren {\frac 1 {2 \sqrt {a^n} } \ln \size {\frac {u - \sqrt {a^n} } {u + \sqrt {a^n} } } } + C\) Primitive of $\dfrac 1 {x^2 - a^2}$
\(\ds \) \(=\) \(\ds \frac 1 {n \sqrt {a^n} } \ln \size {\frac {\sqrt {x^n + a^n} - \sqrt {a^n} } {\sqrt {x^n + a^n} + \sqrt {a^n} } } + C\) substituting for $u$

$\blacksquare$


Sources

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