5.3 The Fundamental Theorem of Calculus/53: Difference between revisions
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<math>= | <math>=\frac{3(9x^2-1)}{9x^2+1}-\frac{2(4x^2-1)}{4x^2+1}</math> | ||
Revision as of 22:21, 6 September 2022
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 g(x)=\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du}
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{d}{dx}g(x) = \frac{d}{dx}\left[\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du\right]=3\cdot\frac{(3x)^2-1}{(3x)^2+1}-2\cdot\frac{(2x)^2-1}{(2x)^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 =(3*\frac{9x^2-1}{9x^2+1})-(2*\frac{4x^2-1}{4x^2+1})}
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Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle ={\frac {3(9x^{2}-1)}{9x^{2}+1}}-{\frac {2(4x^{2}-1)}{4x^{2}+1}}}