f ′ ( x ) = ∫ 1 2 x 4 ( ln ( x ) ) 2 ⋅ d x {\displaystyle f'(x)=\int _{1}^{2}x^{4}(\ln(x))^{2}\cdot dx}
∫ 1 2 x 4 ( ln ( x ) ) 2 ⋅ d x = x 5 ( l n ( x ) ) 2 5 − 2 5 ∫ 1 2 x 4 ln ( x ) ⋅ d x = x 5 ( l n ( x ) ) 2 5 − 2 x 5 ( ln ( x ) ) 2 25 + ∫ 1 2 x 4 ⋅ d x = x 5 ( l n ( x ) ) 2 5 − 2 x 5 ( l n ( x ) ) 2 25 + 2 x 5 25 | 1 2 {\displaystyle \int _{1}^{2}x^{4}(\ln(x))^{2}\cdot dx~~~=~~~{\frac {x^{5}(ln(x))^{2}}{5}}-{\frac {2}{5}}\int _{1}^{2}x^{4}\ln(x)\cdot dx~~~=~~~{\frac {x^{5}(ln(x))^{2}}{5}}-{\frac {2x^{5}(\ln(x))^{2}}{25}}+\int _{1}^{2}x^{4}\cdot dx~~~=~~~{\frac {x^{5}(ln(x))^{2}}{5}}-{\frac {2x^{5}(ln(x))^{2}}{25}}+{\frac {2x^{5}}{25}}{\Bigg |}_{1}^{2}} u = ( ln ( x ) ) 2 d v = d x u = ln ( x ) d v = x 4 {\displaystyle u=(\ln(x))^{2}~~~~~dv=dx~~~~~~~~~~~~~~~u=\ln(x)~~~~~dv=x^{4}} d u = 2 x ln ( x ) ⋅ d x v = x d u = 1 x v = x 5 5 {\displaystyle du={\frac {2}{x}}\ln(x)\cdot dx~~v=x~~~~~~~~~~~du={\frac {1}{x}}~~~~~v={\frac {x^{5}}{5}}} = 32 5 ( ln ( 2 ) ) 2 − 64 25 ( ln ( 2 ) ) + 62 125 {\displaystyle ={\frac {32}{5}}(\ln(2))^{2}-{\frac {64}{25}}(\ln(2))+{\frac {62}{125}}}
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