Primitive of Exponential of a x by Logarithm of x

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Theorem

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


Proof

With a view to expressing the primitive in the form:

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

let:

\(\displaystyle u\) \(=\) \(\displaystyle \ln x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\d u} {\d x}\) \(=\) \(\displaystyle \frac 1 x\) 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 $e^{a x}$


Then:

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

$\blacksquare$


Also see


Sources