∫ 1 2 ( ln x ) 2 x 3 {\displaystyle \int _{1}^{2}{\frac {(\ln {x})^{2}}{x^{3}}}}
u = ln 2 x d v = 1 x 3 {\displaystyle u=\ln ^{2}{x}\qquad dv={\frac {1}{x^{3}}}} d u = 2 ln ( x ) x v = − 1 2 x 2 {\displaystyle du={\frac {2\ln {(x)}}{x}}\qquad v=-{\frac {1}{2x^{2}}}}
∫ 1 2 ln x 2 x 3 = − ln 2 ( x ) 2 x 2 − ∫ − ln ( x ) x 3 = − ln 2 ( x ) 2 x 2 + ∫ ln ( x ) x 3 = − ln 2 ( x ) 2 x 2 − ln ( x ) 2 x 2 − ∫ − 1 2 x 3 u = ln ( x ) d v = 1 x 3 d u = 1 x v = − 1 2 x 2 = − ln 2 ( x ) 2 x 2 − ln ( x ) 2 x 2 − 1 2 ∫ 1 x 3 = − ln ( x ) 2 x 2 − 1 4 x 2 {\displaystyle {\begin{aligned}\int _{1}^{2}{\frac {\ln {x}^{2}}{x^{3}}}=-{\frac {\ln ^{2}{(x)}}{2x^{2}}}-\int -{\frac {\ln {(x)}}{x^{3}}}=-{\frac {\ln ^{2}{(x)}}{2x^{2}}}+\int {\frac {\ln {(x)}}{x^{3}}}&=-{\frac {\ln ^{2}{(x)}}{2x^{2}}}-{\frac {\ln {(x)}}{2x^{2}}}-\int -{\frac {1}{2x^{3}}}\\[2ex]&u=\ln {(x)}\qquad dv={\frac {1}{x^{3}}}\\[2ex]&du={\frac {1}{x}}\qquad v=-{\frac {1}{2x^{2}}}\\[2ex]&=-{\frac {\ln ^{2}{(x)}}{2x^{2}}}-{\frac {\ln {(x)}}{2x^{2}}}-{\frac {1}{2}}\int {\frac {1}{x^{3}}}\\[2ex]&=-{\frac {\ln {(x)}}{2x^{2}}}-{\frac {1}{4x^{2}}}\end{aligned}}}
![{\displaystyle {\begin{aligned}\int _{1}^{2}{\frac {\ln {x}^{2}}{x^{3}}}=-{\frac {\ln ^{2}{(x)}}{2x^{2}}}-\int -{\frac {\ln {(x)}}{x^{3}}}=-{\frac {\ln ^{2}{(x)}}{2x^{2}}}+\int {\frac {\ln {(x)}}{x^{3}}}&=-{\frac {\ln ^{2}{(x)}}{2x^{2}}}-{\frac {\ln {(x)}}{2x^{2}}}-\int -{\frac {1}{2x^{3}}}\\[2ex]&u=\ln {(x)}\qquad dv={\frac {1}{x^{3}}}\\[2ex]&du={\frac {1}{x}}\qquad v=-{\frac {1}{2x^{2}}}\\[2ex]&=-{\frac {\ln ^{2}{(x)}}{2x^{2}}}-{\frac {\ln {(x)}}{2x^{2}}}-{\frac {1}{2}}\int {\frac {1}{x^{3}}}\\[2ex]&=-{\frac {\ln {(x)}}{2x^{2}}}-{\frac {1}{4x^{2}}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/5ba150cd782eeb8763f45ce588aa2e82692d6239)