Primitive of Exponential of a x by Logarithm of x

From ProofWiki
Jump to navigation Jump to search


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


With a view to expressing the primitive in the form:

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


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

and let:

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


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


Also see