Задача №1414
Условие
Найти интеграл \(\int\frac{\ln^3{x}dx}{\sqrt{x^5}}\)
Решение
\[
\int\frac{\ln^3{x}dx}{\sqrt{x^5}}
=\int{x^{-\frac{5}{2}}}\ln^3{x}dx
=\left[\begin{aligned}
& u=\ln^3{x};\;du=\frac{3\ln^2{x}dx}{x}.\\
& dv=x^{-\frac{5}{2}};\;v=-\frac{2}{3}x^{-\frac{3}{2}}.
\end{aligned}\right]
=-\frac{2\ln^3{x}}{3\sqrt{x^3}}+2\cdot\int{x^{-\frac{5}{2}}}\ln^2{x}dx=\\
=\left[\begin{aligned}
& u=\ln^2{x};\;du=\frac{2\ln{x}dx}{x}.\\
& dv=x^{-\frac{5}{2}};\;v=-\frac{2}{3}x^{-\frac{3}{2}}.
\end{aligned}\right]
=-\frac{2\ln^3{x}}{3\sqrt{x^3}}-\frac{4\ln^2{x}}{3\sqrt{x^3}}+\frac{8}{3}\cdot\int{x^{-\frac{5}{2}}}\ln{x}dx=\\
=\left[\begin{aligned}
& u=\ln{x};\;du=\frac{dx}{x}.\\
& dv=x^{-\frac{5}{2}};\;v=-\frac{2}{3}x^{-\frac{3}{2}}.
\end{aligned}\right]
=-\frac{2\ln^3{x}}{3\sqrt{x^3}}-\frac{4\ln^2{x}}{3\sqrt{x^3}}-\frac{16\ln{x}}{9\sqrt{x^3}}+\frac{16}{9}\cdot\int{x^{-\frac{5}{2}}}dx=\\
=-\frac{2\ln^3{x}}{3\sqrt{x^3}}-\frac{4\ln^2{x}}{3\sqrt{x^3}}-\frac{16\ln{x}}{9\sqrt{x^3}}-\frac{32}{27\sqrt{x^3}}+C
=-\frac{2}{3x\sqrt{x}}\cdot\left(\ln^3{x}+2\ln^2{x}+\frac{8\ln{x}}{3}+\frac{16}{9}\right)+C
\]
Ответ:
\(-\frac{2}{3x\sqrt{x}}\cdot\left(\ln^3{x}+2\ln^2{x}+\frac{8\ln{x}}{3}+\frac{16}{9}\right)+C\)