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>\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{1}{x^2}</math> | ||
<math>\left[(3)\frac{(3x)^2-1}{(3x)^2+1}du\right] | |||
Revision as of 15:57, 26 August 2022
![{\displaystyle {\frac {d}{dx}}\left[\int _{2x}^{3x}{\frac {u^{2}-1}{u^{2}+1}}du\right]=3*{\frac {1}{x^{2}}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/5e9d28835ec665c9819d17ee8857a9f780e294a7)
<math>\left[(3)\frac{(3x)^2-1}{(3x)^2+1}du\right]