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 {\displaystyle \int _{1}^{5}\left(5\ln(x)\right)dx=5\int _{1}^{5}\left(\ln(x)\right)dx}
Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} u = \ln(x) \qquad dv= 1 dx \ du = \frac{1}{x} dx \quad v=x \\ \end{align} }