y = 5 ln ( x ) , y = x ln ( x ) {\displaystyle y={\color {OliveGreen}5\ln(x)},y={\color {Red}x\ln(x)}}
5 ln ( x ) = x ln ( x ) x = 5 x = 1 {\displaystyle {\begin{aligned}5\ln(x)&=x\ln(x)\\[1ex]&x=5\\[1ex]&x=1\\[1ex]\end{aligned}}}
When x = 2 {\displaystyle x=2} , 5 ln ( 2 ) > 2 ln ( 2 ) {\displaystyle 5\ln(2)>2\ln(2)}
∫ 1 5 ( 5 ln ( x ) − x ln ( x ) ) d x = ∫ 1 5 ( 5 ln ( x ) ) d x − ∫ 1 5 ( x ln ( x ) ) d x = 25 ln ( 5 ) − 20 − ( 25 2 ln ( 5 ) − 6 ) = 25 2 ln ( 5 ) − 14 {\displaystyle \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}
∫ 1 5 ( 5 ln ( x ) ) d x = 5 ∫ 1 5 ( ln ( x ) ) d x = 5 ( x ln ( x ) | 1 5 − ∫ 1 5 ( x x ) d x ) = 5 ( x ln ( x ) | 1 5 − x | 1 5 ) = 5 ( 5 ln ( 5 ) − 1 ln ( 1 ) − ( 5 − 1 ) ) = 25 ln ( 5 ) − 20 u = ln ( x ) d v = 1 d x d u = 1 x d x v = x {\displaystyle {\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}}}
∫ 1 5 ( x ln ( x ) ) d x = x 2 ln ( x ) 2 | 1 5 − ∫ 1 5 ( x 2 2 x ) d x = x 2 ln ( x ) 2 | 1 5 − 1 2 ∫ 1 5 ( x ) d x = 1 ln ( 1 ) 2 − 25 ln ( 5 ) 2 − ( 1 2 ) ( x 2 2 ) | 1 5 = 0 − 25 2 ln ( 5 ) − ( 25 − 1 4 ) = 25 2 ln ( 5 ) − 6 {\displaystyle {\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)-\left({\frac {25-1}{4}}\right)={\frac {25}{2}}\ln(5)-6}
u = ln ( x ) d v = x d x d u = 1 x v = x 2 2 {\displaystyle {\begin{aligned}u&=\ln(x)\quad dv=xdx\\du&={\frac {1}{x}}\quad v={\frac {x^{2}}{2}}\\\end{aligned}}}
![{\displaystyle {\begin{aligned}5\ln(x)&=x\ln(x)\\[1ex]&x=5\\[1ex]&x=1\\[1ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/7ecb8b4109c7b052ff090a770a426d465b1ce8f1)
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![{\displaystyle {\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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/25d25c0f3562f14a2719a35ea2cea6a5b8a0b4b0)
