7.1 Integration By Parts/54

$y=5\ln(x),y=x\ln(x)$

${\begin{aligned}5\ln(x)&=x\ln(x)\\[1ex]&x=5\\[1ex]&x=1\\[1ex]5\ln(2)>2\ln(2)\end{aligned}}$

$\int _{1}^{5}\left(5\ln(x)-x\ln(x)\right)dx=\int _{1}^{5}\left(5\ln(x)\right)dx-\int _{1}^{5}\left(x\ln(x)\right)dx$

$\int _{1}^{5}\left(5\ln(x)\right)dx=5\int _{1}^{5}\left(\ln(x)\right)dx=5\left(x\ln(x){\bigg |}_{1}^{5}-\int _{1}^{5}\left({\frac {x}{x}}\right)dx\right)=5\left(x\ln(x){\bigg |}_{1}^{5}-x{\bigg |}_{1}^{5}\right)=5\left(5\ln(5)-1\ln(1)-\left(5-1\right)\right)=25\ln(5)-20$

${\begin{aligned}u&=\ln(x)\quad dv=1dx\\du&={\frac {1}{x}}dx\quad v=x\\\end{aligned}}$

Failed to parse (syntax error): {\displaystyle \int_{1}^{5} \left(x\ln(x) \right)dx = \frac{x^2\ln(x)}{2}\bigg|_{1}^{5} - \int_{1}^{5} \left(\frac{x^2}{2x} \right)dx = \frac{x^2\ln(x)}{2}\bigg|_{1}^{5} - \frac{1}{2}\int_{1}^{5} \left(x \right)dx = 1\ln(1)-25\ln(5) -\left(\frac{1}{2})\right \left( \frac{x^2}{2} \right) \bigg|_{1}^{5} }

${\begin{aligned}u&=\ln(x)\quad dv=xdx\\du&={\frac {1}{x}}\quad v={\frac {x^{2}}{2}}\\\end{aligned}}$