Answer
$$\int\frac{(\ln x)^{-3}}{x}dx=-\frac{1}{2(\ln x)^2}+C$$
Work Step by Step
$$A=\int\frac{(\ln x)^{-3}}{x}dx$$
We set $a=\ln x$, which means $$da=\frac{1}{x}dx$$
Therefore, $$A=\int a^{-3}da$$ $$A=\frac{a^{-2}}{-2}+C$$ $$A=-\frac{1}{2a^2}+C$$ $$A=-\frac{1}{2(\ln x)^2}+C$$
