7.1 Integration By Parts/31

$f'(x)=\int _{1}^{2}x^{4}(\ln(x))^{2}\cdot dx$

$\int _{1}^{2}x^{4}(\ln(x))^{2}\cdot dx~~~=~~~{\frac {x^{5}(ln(x))^{2}}{5}}-{\frac {2}{5}}\int _{1}^{2}x^{4}\ln(x)\cdot dx~~~=~~~{\frac {x^{5}(ln(x))^{2}}{5}}-{\frac {2x^{5}(\ln(x))^{2}}{25}}+\int _{1}^{2}x^{4}\cdot dx~~~=~~~{\frac {x^{5}(ln(x))^{2}}{5}}-{\frac {2x^{5}(ln(x))^{2}}{25}}+{\frac {2x^{5}}{25}}{\Bigg |}_{1}^{2}$
$u=(\ln(x))^{2}~~~~~dv=dx~~~~~~~~~~~~~~~u=\ln(x)~~~~~dv=x^{4}$
$du={\frac {2}{x}}\ln(x)\cdot dx~~v=x~~~~~~~~~~~du={\frac {1}{x}}~~~~~v={\frac {x^{5}}{5}}$

$={\frac {32}{5}}(\ln(2))^{2}-{\frac {64}{25}}(\ln(2))+{\frac {62}{125}}$

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