5.3 The Fundamental Theorem of Calculus/1: Difference between revisions

Revision as of 19:28, 23 August 2022

$\int {\frac {x}{\sqrt {x^{2}+1}}}dx={\sqrt {x^{2}+1}}+c$

${\frac {d}{dx}}\left[(x^{2}+1)^{\frac {1}{2}}+c\right]$

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	<math>\int\frac{x}{\sqrt{x^2+1}}dx=\sqrt{x^2+1}+c</math>		<math>\int\frac{x}{\sqrt{x^2+1}}dx=\sqrt{x^2+1}+c</math>

	<math>~~\left[~~\frac{d}{dx}(x^2+1)^\frac{1}{2}+c\right]</math>		<math>\frac{d}{dx}\left[(x^2+1)^\frac{1}{2}+c\right]</math>