How do I evaluate $\int \log(x) e^x\;dx ?$ I tried to do integration by parts...
$$\int\log(x) \; dx = (x-1)\log(x) $$
Let $I=\int\log(x) e^x \; dx$. Therefore, $$ I = (x-1)\log(x) e^x - \int (x-1)\log(x) e^x \; dx $$ $$= (x-1)\log(x) e^x - \int (x)\log(x) e^x \; dx +\int \log(x) e^x \; dx$$ $$= (x-1)\log(x) e^x - \int x \log(x) e^x \; dx + I$$
Now what to do, $I$ is on both LHS and RHS??