7.1 Integration By Parts/26

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Revision as of 06:19, 29 November 2022 by Jorgen70034@students.laalliance.org (talk | contribs) (Created page with "<math>\int_{1}^{\sqrt{3}}arctan\left(\frac{1}{x}\right)dx</math> <math>letu =arctan\left(\frac{1}{x}\right)du=dx=du=\frac{1}{1+(1/x)^2}x\frac{-1}{x^2}dx=\frac{-dx}{x^2+1}</math> <math>\int_{1}^{\sqrt{3}}arctan\frac{1}{x}dx=[xarctan\frac{1}{x}]\bigg|_{0}^{1}+\int_{1}^{\sqrt{3}}\frac{x}{dx}x^2+1=\sqrt{3}\frac{\pi}{6}=\frac{1}{4}=\frac{1}{2}[in(x^2+1)]\bigg|_{1}^{\sqrt{3}}</math> <math>=\frac{\pi\sqrt{3}}{6}-\frac{\pi}{4}(in4-in2)=\frac{\pi\sqrt{3}}{6}-\frac{\pi}{2}=\fr...")
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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}^{\sqrt{3}}arctan\left(\frac{1}{x}\right)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 letu =arctan\left(\frac{1}{x}\right)du=dx=du=\frac{1}{1+(1/x)^2}x\frac{-1}{x^2}dx=\frac{-dx}{x^2+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 \int_{1}^{\sqrt{3}}arctan\frac{1}{x}dx=[xarctan\frac{1}{x}]\bigg|_{0}^{1}+\int_{1}^{\sqrt{3}}\frac{x}{dx}x^2+1=\sqrt{3}\frac{\pi}{6}=\frac{1}{4}=\frac{1}{2}[in(x^2+1)]\bigg|_{1}^{\sqrt{3}}}

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{\pi\sqrt{3}}{6}-\frac{\pi}{4}(in4-in2)=\frac{\pi\sqrt{3}}{6}-\frac{\pi}{2}=\frac{1}{2}in\frac{4}{2}=\frac{\pi\sqrt{3}}{6}-\frac{\pi}{2}+\frac{1}{2}in2}

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