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 = \int_{1}^{5} \left(5\ln(x) \right)dx - \int_{1}^{5} \left(x\ln(x) \right)dx }

$\int _{1}^{5}\left(5\ln(x)\right)dx=5\int _{1}^{5}\left(\ln(x)\right)dx=5\left(x\ln(x){\bigg |}_{1}^{5}-\int _{1}^{5}\left({\frac {x}{x}}\right)dx\right)=5\left(x\ln(x){\bigg |}_{1}^{5}-x{\bigg |}_{1}^{5}\right)=5\left(5\ln(5)-1\ln(1)-\left(5-1\right)\right)=25\ln(5)-20$

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} u &= \ln(x) \quad dv= 1 dx \\ du &= \frac{1}{x} dx \quad v=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 \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\\ u &= \ln(x) \quad dv= x dx \\ du &= \frac{1}{x} \quad v=\frac{x^2}{2} \\ \end{align} }