∫ 1 2 ( ln x ) 2 x 3 {\displaystyle \int _{1}^{2}{\frac {(\ln {x})^{2}}{x^{3}}}}
u = ln x d v = 1 x 3 {\displaystyle u=\ln {x}\qquad dv={\frac {1}{x^{3}}}} d u = 1 x v = − 1 2 x 2 {\displaystyle du={\frac {1}{x}}\qquad v=-{\frac {1}{2x^{2}}}}
∫ 1 2 ln x 2 x 3 = − ln 2 ( x ) 2 x 2 − ∫ − 1 2 x 3 = − 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 {1}{2x^{3}}}\\[2ex]&=-{\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 {1}{2x^{3}}}\\[2ex]&=-{\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/401b5b4aef7cbc4fb7a91dad8a58bdbb4317ccc7)