7.1 Integration By Parts/28

$\int _{1}^{2}{\frac {(\ln {x})^{2}}{x^{3}}}$

$u=\ln ^{2}{x}\qquad dv={\frac {1}{x^{3}}}$

$du={\frac {2\ln {(x)}}{x}}\qquad v=-{\frac {1}{2x^{2}}}$

${\begin{aligned}\int _{1}^{2}{\frac {\ln {x}^{2}}{x^{3}}}=-{\frac {\ln ^{2}{(x)}}{2x^{2}}}-\int -{\frac {\ln {(x)}}{x^{3}}}&=-{\frac {\ln ^{2}{(x)}}{2x^{2}}}+\int {\frac {\ln {(x)}}{x^{3}}}\\[2ex]&u=\ln {(x)}\qquad dv={\frac {1}{x^{3}}}\\[2ex]&du={\frac {1}{x}}\qquad v=-{\frac {1}{2x^{2}}}\\[2ex]&=-{\frac {\ln ^{2}{(x)}}{2x^{2}}}-{\frac {\ln {(x)}}{2x^{2}}}-\int -{\frac {1}{2x^{3}}}\\[2ex]&=-{\frac {\ln ^{2}{(x)}}{2x^{2}}}-{\frac {\ln {(x)}}{2x^{2}}}-{\frac {1}{2}}\int {\frac {1}{x^{3}}}\\[2ex]&=-{\frac {\ln ^{2}{(x)}}{2x^{2}}}-{\frac {\ln {(x)}}{2x^{2}}}-{\frac {1}{4x^{2}}}{\bigg |}_{1}^{2}\\[2ex]&=-{\frac {\ln ^{2}{(2)}}{8}}-{\frac {\ln {(2)}}{8}}-{\frac {3}{16}}\\[2ex]\end{aligned}}$