5.3 The Fundamental Theorem of Calculus/33: Difference between revisions

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<math>\frac{d}{dx}\left[h(x)\right]=\frac{d}{dx}\left[int_{2}^{1/x}\arctan(t)dt\right]
<math>\frac{d}{dx}\left[h(x)\right]=\frac{d}{dx}\left[int_{2}^{1/x}\arctan(t)dt\right]


\frac{-1}{x^2}\left(\arctan(\frac{1}{x}\right)-0</math>
<math>\frac{-1}{x^2}\left(\arctan(\frac{1}{x}\right)-0</math>

Revision as of 19:50, 25 August 2022