Primitive of Exponential of a x by Logarithm of x

Theorem

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

Proof
With a view to expressing the primitive in the form:
 * $\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$

let:

and let:

Then:

Also see

 * Primitive of $\dfrac {e^{a x} } x$