integrating differential form using integration by parts

Question

I would like to integrate $$-xdlog(y(x))$$ with respect to $x$, where $d$ is the exterior derivative. I'm trying to use integration by parts, so $$-\int xdlog(y(x)) \; dx $$ letting \begin{align*} u&=x , \ \ \ dv=dlog(y(x)) dx \\ du&=dx , \ \ v= ? \end{align*} I'm getting stuck on how to write $v$. Is there a better way to integrate this?

Answer 1

If you are working with differential forms, then $d(\log y)$ is already a differential one-form given by

$$ d(\log y) = \frac{y'(x)}{y(x)} \, dx $$

so there's no need for the extra $d$. Then, you have $u = x$, $v = \log(y)$ and

$$ -\int x \frac{y'(x)}{y(x)} \, dx = \int x d(\log y) = -\int u dv = \int v du - uv = \int \log(y(x)) \, dx - x \log(y(x)) + C. $$