5.3 The Fundamental Theorem of Calculus/53: Difference between revisions
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<math>\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du</math> | <math>g(x)=\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du</math> | ||
<math>\frac{d}{dx}\left[\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du\right]=(3*\frac{3x^2-1}{3x^2+1})-(2*\frac{2x^2-1}{2x^2+1})</math> | <math>\frac{d}{dx}g(x) = \frac{d}{dx}\left[\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du\right]=(3*\frac{3x^2-1}{3x^2+1})-(2*\frac{2x^2-1}{2x^2+1})</math> | ||
Revision as of 22:19, 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*\frac{3x^2-1}{3x^2+1})-(2*\frac{2x^2-1}{2x^2+1})}
=
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}})}