Primitive of Exponential of a x by Logarithm of x

From ProofWiki
Jump to navigation Jump to search

Theorem

$\ds \int e^{a x} \ln x \rd x = \frac {e^{a x} \ln x} a - \frac 1 a \int \frac {e^{a x} } x \rd x + C$


Proof

With a view to expressing the primitive in the form:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

\(\ds u\) \(=\) \(\ds \ln x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds \frac 1 x\) Derivative of Natural Logarithm


and let:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds e^{a x}\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \frac {e^{a x} } a\) Primitive of $e^{a x}$


Then:

\(\ds \int e^{a x} \ln x \rd x\) \(=\) \(\ds \ln x \paren {\frac {e^{a x} } a} - \int \paren {\frac {e^{a x} } a} \frac 1 x \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac {e^{a x} \ln x} a - \frac 1 a \int \frac {e^{a x} } x \rd x + C\) Primitive of Constant Multiple of Function

$\blacksquare$


Also see


Sources