7.1 Integration By Parts/54

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

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

When $x=2$, $5\ln(2)>2\ln(2)$

$\int _{1}^{5}\left(5\ln(x)-x\ln(x)\right)dx={\color {NavyBlue}\int _{1}^{5}\left(5\ln(x)\right)dx}-{\color {RedOrange}\int _{1}^{5}\left(x\ln(x)\right)dx}=25\ln(5)-20-\left({\frac {25}{2}}\ln(5)-6\right)={\frac {25}{2}}\ln(5)-14$

${\begin{aligned}{\color {NavyBlue}\int _{1}^{5}\left(5\ln(x)\right)dx}&=5\int _{1}^{5}\left(\ln(x)\right)dx=5\left(x\ln(x){\bigg |}_{1}^{5}-\int _{1}^{5}\left({\frac {x}{x}}\right)dx\right)=5\left(x\ln(x){\bigg |}_{1}^{5}-x{\bigg |}_{1}^{5}\right)=5\left(5\ln(5)-1\ln(1)-\left(5-1\right)\right)=25\ln(5)-20\\[2ex]u&=\ln(x)\quad dv=1dx\\[2ex]du&={\frac {1}{x}}dx\quad v=x\\[2ex]\end{aligned}}$

${\color {RedOrange}\int _{1}^{5}\left(x\ln(x)\right)dx}={\frac {x^{2}\ln(x)}{2}}{\bigg |}_{1}^{5}-\int _{1}^{5}\left({\frac {x^{2}}{2x}}\right)dx={\frac {x^{2}\ln(x)}{2}}{\bigg |}_{1}^{5}-{\frac {1}{2}}\int _{1}^{5}\left(x\right)dx={\frac {1\ln(1)}{2}}-{\frac {25\ln(5)}{2}}-\left({\frac {1}{2}}\right)\left({\frac {x^{2}}{2}}\right){\bigg |}_{1}^{5}=0-{\frac {25}{2}}\ln(5)-{\frac {1}{2}}\left({\frac {25-1}{2}}\right)={\frac {25}{2}}\ln(5)-6$

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