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

Revision as of 19:31, 23 August 2022

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

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

let $a=x^{2}+1$ and $b=a^{1/2}$ then ${\frac {da}{dx}}=2x{\text{and}}{\frac {db}{da}}={\frac {1}{2}}*a^{-1/2}$

Revision as of 19:29, 23 August 2022 (view source) Dvaezazizi@laalliance.org (talk \| contribs) No edit summary ← Older edit		Revision as of 19:31, 23 August 2022 (view source) Dvaezazizi@laalliance.org (talk \| contribs) No edit summary Newer edit →
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	<math>\frac{d}{dx}\left[(x^2+1)^\frac{1}{2}+c\right]</math>		<math>\frac{d}{dx}\left[(x^2+1)^\frac{1}{2}+c\right]</math>

	let <math>a=x^2+1</math> and <math>b=a^{1/2}</math>		let <math>a=x^2+1</math> and <math>b=a^{1/2}</math> then <math>\frac{da}{dx}=2x \text{and} \frac{db}{da}=\frac{1}{2}*a^{-1/2}</math>