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{3x^2-1}{3x^2+1}</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})</math> | ||
Revision as of 15:59, 26 August 2022
![{\displaystyle {\frac {d}{dx}}\left[\int _{2x}^{3x}{\frac {u^{2}-1}{u^{2}+1}}du\right]=(3*{\frac {3x^{2}-1}{3x^{2}+1}})}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/602b414af3b3f948d46438f111f8c8bbaa8276fb)