Задача №1413
Условие
Найти интеграл \(\int\frac{\ln^3x}{x^2}dx\).
Решение
\[
\int\frac{\ln^3x}{x^2}dx
=\left[\begin{aligned}& u=\ln^3x;\,du=\frac{3\ln^2{x}dx}{x}.\\& dv=\frac{dx}{x^2};\;v=-\frac{1}{x}.\end{aligned}\right]
=-\frac{\ln^3x}{x}+3\int\frac{\ln^2x}{x^2}dx
=\left[\begin{aligned}& u=\ln^2x;\,du=\frac{2\ln{x}dx}{x}.\\& dv=\frac{dx}{x^2};\;v=-\frac{1}{x}.\end{aligned}\right]=\\
=-\frac{\ln^3x}{x}+3\cdot\left(-\frac{\ln^2x}{x}+2\int\frac{\ln{x}}{x^2}dx\right)
=-\frac{\ln^3x}{x}-\frac{3\ln^2x}{x}+6\int\frac{\ln{x}}{x^2}dx
=\left[\begin{aligned}& u=\ln{x};\,du=\frac{dx}{x}.\\& dv=\frac{dx}{x^2};\;v=-\frac{1}{x}.\end{aligned}\right]=\\
=-\frac{\ln^3x}{x}-\frac{3\ln^2x}{x}+6\cdot\left(-\frac{\ln{x}}{x}+\int\frac{dx}{x^2}\right)
=-\frac{\ln^3x}{x}-\frac{3\ln^2x}{x}+6\cdot\left(-\frac{\ln{x}}{x}-\frac{1}{x}\right)+C=\\
=-\frac{\ln^3x}{x}-\frac{3\ln^2x}{x}-\frac{6\ln{x}}{x}-\frac{6}{x}+C
=\frac{-\ln^3x-3\ln^2x-6\ln{x}-6}{x}+C
\]
Ответ:
\(\frac{-\ln^3x-3\ln^2x-6\ln{x}-6}{x}+C\)