How to find $\int \frac{\ln(x)}{x^2}dx$ I need to find
$$\int \frac{\ln(x)}{x^2}dx.$$
I have tried substitution with $u=\ln(x)$, then $du = 1/x dx$, but this only takes care of one of the $x$ on the bottom:
$$
\int \frac{u}{x} du.
$$
I could obviously replace the $x$ with $e^u$, but then I get
$$
\int u e^{-u} du.
$$
Then one could do integration by parts.
I am pretty sure that there has to be an easier way to do this one. Therefore my question is: how can one do this in a different way?
 A: Another way : $$\begin{align}\int\frac{\ln x}{x^2}dx&=\int\left(-\frac 1x\right)'\ln xdx\\&=-\frac{\ln x}{x}-\int \left(-\frac 1x\right)\cdot \frac 1xdx\\&=-\frac{\ln x}{x}-\frac1x +C\end{align}$$
A: Let's integrate by parts setting $u=\log{x}$ and $dv=\frac{dx}{x^2}=-d\left(\frac{1}{x}\right)$
We have
$$\int\frac{\log{x}}{x^2}dx=-\frac{\log{x}}{x}+\int\frac{1}{x^2}dx=-\frac{1+\log{x}}{x}+K$$
A: Another way $$\int\frac{\log\left(x\right)}{x^{2}}dx=\frac{\log\left(x\right)}{x}-\int x\left(\frac{1}{x^{3}}-2\frac{\log\left(x\right)}{x^{3}}\right)dx=\frac{\log\left(x\right)}{x}+\frac{1}{x}+2\int\frac{\log\left(x\right)}{x^{2}}dx
 $$ and so $$\int\frac{\log\left(x\right)}{x^{2}}dx=-\frac{\log\left(x\right)+1}{x}+C.
 $$
A: By inspection, the integrand is a term in the derivative of $-\dfrac{\ln(x)}x$, and the other term would be $-\dfrac1{x^2}$.
Hence
$$-\frac{\ln(x)}x-\frac1x+C.$$

Or, by inspection, $\dfrac{dx}{x^2}$ hints a substitution with $\dfrac1x$, resulting in the integrand $-\ln(x)$, that you find in a table, and
$$\frac1x\ln\left(\frac1x\right)-\frac1x+C.$$
