y = 5 ln ( x ) , y = x ln ( x ) {\displaystyle y=5\ln(x),y=x\ln(x)}
5 ln ( x ) = x ln ( x ) x = 5 x = 1 5 ln ( 2 ) > 2 ln ( 2 ) {\displaystyle {\begin{aligned}5\ln(x)&=x\ln(x)\\[1ex]&x=5\\[1ex]&x=1\\[1ex]5\ln(2)>2\ln(2)\end{aligned}}}
∫ 1 5 ( 5 ln ( x ) − x ln ( x ) ) d x = ∫ 1 5 ( 5 ln ( x ) ) d x − ∫ 1 5 ( x ln ( x ) ) d x {\displaystyle \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}
∫ 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 ) {\displaystyle \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)}
u = ln ( x ) d v = 1 d x d u = 1 x d x v = x {\displaystyle {\begin{aligned}u&=\ln(x)\quad dv=1dx\\du&={\frac {1}{x}}dx\quad v=x\\\end{aligned}}}