how to find the integral of a rational logarithmic function

Question

I can't seem to figure this one out,

it is:

$$\int\frac{\ln(x)}xdx $$

I substituted $u$ for $\ln(x)$, so $u = \ln(x)$ and $du = \frac1x dx$

then to find $x$ in terms of $u$: $e^u = x$

so I get

$$\int\frac{u}{e^{u^2}}du$$

from here I can't figure out where to go, I have tried playing around with the numbers but after a few hours I figured I'd ask someone here.

I sense that I must somehow get it to the form $$\int\frac1xdx,$$ but i an not sure how to get a $1$ in the numerator.

Answer 1

Your replacement of variables is wrong. Since $u=\ln x$, you get $$\int\ln x \cdot \frac{1}{x}dx$$ in which you replace $\ln x$ with $u$ and $\frac1x dx$ with $du$. What you get does not contain any $e^u$ or $e^{u^2}$.

Answer 2

Notice that $\frac 1 x$ is the derivation of $\log x$ so your integral has the form $$\int f(x)f'(x)dx$$ can you take it from here?