🔢Para resolver $\displaystyle \int \frac{ln^2(x)}{x^2}\, dx$ $\quad$ usaremos la técnica de integración por partes, donde

$$u={\ln }^{2}x,dv=\frac{dx}{x^2}.$$ ✏️ Cálculo de $du$ y $v$

$$\begin{array}{rl}du& =2\ln x\frac{dx}{x}=\frac{2\ln x}{x}dx,\\ v& =\int x^{-2}dx=-x^{-1}=-\frac{1}{x}.\end{array}$$

✏️ Aplicación de la fórmula

$$\int udv=uv-\int vdu$$

$$\begin{array}{rl}\int \frac{{\ln }^{2}x}{x^2}dx& =uv-\int vdu\\ & =({\ln }^{2}x)(-\frac{1}{x})-\int (-\frac{1}{x})\frac{2\ln x}{x}dx\\ & =-\frac{{\ln }^{2}x}{x}+2\int \frac{\ln x}{x^2}dx.\end{array}$$

✏️ Integrar ${\displaystyle \int \frac{\ln x}{x^2}dx}$ $\quad$ nuevamente aplicación de partes con

$$\tilde{u}=\ln x,d\tilde{v}=\frac{dx}{x^2}.$$

$$\begin{array}{rl}d\tilde{u}& =\frac{dx}{x},\tilde{v}=-\frac{1}{x}.\end{array}$$

$$\begin{array}{rl}\int \frac{\ln x}{x^2}dx& =\tilde{u}\tilde{v}-\int \tilde{v}d\tilde{u}\\ & =\ln x(-\frac{1}{x})-\int (-\frac{1}{x})\frac{dx}{x}\\ & =-\frac{\ln x}{x}+\int \frac{dx}{x^2}\\ & =-\frac{\ln x}{x}-\frac{1}{x}+C.\end{array}$$

Multiplicamos por 2 (el factor que estaba fuera de la integral):

$$2\int \frac{\ln x}{x^2}dx=-\frac{2\ln x}{x}-\frac{2}{x}.$$

✏️ Resultado final

$$\int \frac{{\ln }^{2}x}{x^2}dx=-\frac{{\ln }^{2}x}{x}-\frac{2\ln x}{x}-\frac{2}{x}+C.$$

Se puede factorizar el denominador:

$$\boxed{{\displaystyle {\displaystyle \int \frac{{\ln }^{2}x}{x^2}dx=-\frac{1}{x}({\ln }^{2}x+2\ln x+2)+C}}}.$$