Answer
$$-\frac{(\ln x)^{2}}{x}-2 \frac{\ln x}{x}-2 \frac{1}{x}+C$$
Work Step by Step
Given $$\int \frac{(\ln x)^{2} d x}{x^{2}}$$
Let $$u =\ln x\ \ \ \ \ \ du =\frac{1}{x}dx $$
Then
$$\int \frac{(\ln x)^{2} d x}{x^{2}}= \int u^{2} e^{-u} d u $$
Use $$\int x^{n} e^{-x} d x=-x^{n} e^{-x}+n \int x^{n-1} e^{-x} d x $$
Then
\begin{aligned}
\int u^{2} e^{-u} d u &=-u^{2} e^{-u}+2 \int u^{2-1} e^{-u} d u \\
&=-u^{2} e^{-u}+2 \int u e^{-u} d u \\
&=-u^{2} e^{-u}-2 u e^{-u}+2 \int e^{-u} d u \\
&=-u^{2} e^{-u}-2 u e^{-u}-2 e^{-u}+C \\
&=-\frac{(\ln x)^{2}}{x}-2 \frac{\ln x}{x}-2 \frac{1}{x}+C
\end{aligned}
