Primitive of Power of x over Logarithm of x

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Theorem

For $x > 1$:

\(\ds \int \frac {x^m \rd x} {\ln x}\) \(=\) \(\ds \map \ln {\ln x} + \sum_{k \mathop \ge 1} \frac {\paren {m + 1}^k \paren {\ln x}^k} {k \times k!} + C\)
\(\ds \) \(=\) \(\ds \map \ln {\ln x} + \dfrac {\paren {m + 1} \ln x} {1 \times 1!} + \frac {\paren {m + 1}^2 \paren {\ln x}^2} {2 \times 2!} + \dfrac {\paren {m + 1}^3 \paren {\ln x}^3} {3 \times 3!} + \cdots + C\)


Proof

\(\ds u\) \(=\) \(\ds \ln x\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds e^u\)
\(\ds \leadsto \ \ \) \(\ds \d x\) \(=\) \(\ds e^u \d u\)
\(\ds \leadsto \ \ \) \(\ds \int \frac {x^m \rd x} {\ln x}\) \(=\) \(\ds \int \frac {\paren {e^u}^m e^u \rd u} u\)
\(\ds \) \(=\) \(\ds \int \frac {e^{u \paren {m + 1} } \rd u} u\)
\(\ds \) \(=\) \(\ds \ln u + \sum_{k \mathop \ge 1} \frac {\paren {\paren {m + 1} u}^k} {k \times k!} + C\) Primitive of $\dfrac {e^{a x} } x$: $u > 0$ for $x > 1$
\(\ds \) \(=\) \(\ds \map \ln {\ln x} + \sum_{k \mathop \ge 1} \frac {\paren {m + 1}^k \paren {\ln x}^k} {k \times k!} + C\) substituting $u = \ln x$

$\blacksquare$


Sources

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