# Primitive of Exponential of a x by Logarithm of x

## 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

$\blacksquare$