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

Revision as of 19:38, 25 August 2022

$h(x)=\int _{2}^{1/x}\arctan(t)dt$

<math>h'(x)=\-frac{1}{x^2}

Revision as of 19:36, 25 August 2022 (view source) Ricardom59381@students.laalliance.org (talk \| contribs) No edit summary ← Older edit		Revision as of 19:38, 25 August 2022 (view source) Ricardom59381@students.laalliance.org (talk \| contribs) No edit summary Newer edit →
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	<math>h(x)=\int_{2}^{1/x}\arctan(t)dt</math>		<math>h(x)=\int_{2}^{1/x}\arctan(t)dt</math>

			<math>h'(x)=\-frac{1}{x^2}