7.1 Integration By Parts/54

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle y=5\ln(x) , y=x\ln(x) }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} 5\ln(x) &=x\ln(x)\\[1ex] &x=5 \\[1ex] &x=1 \\[1ex] 5\ln(2) > 2\ln(2) \end{align} }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\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 }

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

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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)-{\frac {1}{2}}\left({\frac {25-1}{2}}\right)={\frac {25}{2}}\ln(5)-6}

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