5.5 The Substitution Rule/59

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}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx }

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 &=\frac{1}{x} \\[2ex] du &=-\frac{1}{x^2}dx \\[2ex] -du &=\frac{1}{x^2}dx \\[2ex] \end{align} }

New upper limit: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 \frac{1}{2} = \frac{1}{(2)} }
New lower limit: 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 1 = \frac{1}{(1)} }

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} \int_{1}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx &=\int_{1}^{2} e^\frac{1}{x}(\frac{1}{x^2}\,dx) &=\int_{1}^{\frac{1}{2}}e^u\,(-du) \\[2ex] &=-\int_{1}^{\frac{1}{2}}e^u\,du \\[2ex] &=-e^u\bigg|_{1}^{\frac{1}{2}} \\[2ex] &=-\sqrt{e} - (-e^1) \\[2ex] &=e-\sqrt{e} \end {align} }