Antiderivative of $e^{au}$ I cannot seem to figure out $\int e^{au}du$.  I have tried u-substitution (of course with a variable other than u) and can't get it to work out to the right answer.  Any help is appreciated.

Thanks!
Notice $$d(e^{au}) = a e^{au} du \implies \int d(e^{au}) = \int a e^{au} du \implies \int e^{au} du = \frac{1}{a} e^{au} + K$$
$$\frac d{du}\left(\frac1ae^{au}\right)=e^{au}$$