5.4 Indefinite Integrals and the Net Change Theorem/1: Difference between revisions

Revision as of 16:29, 25 August 2022

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

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

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

${\frac {da}{dx}}{\frac {db}{da}}=\left(2x\right)\left({\frac {1}{2}}a^{-1/2}\right)=xa^{-1/2}=x(x^{2}+1)^{-1/2}={\frac {x}{\sqrt {x^{2}+1}}}$

Revision as of 20:02, 23 August 2022 (view source) Dvaezazizi@laalliance.org (talk \| contribs) (Created page with "<math>\int\frac{x}{\sqrt{x^2+1}}dx=\sqrt{x^2+1}+c</math> <math>\frac{d}{dx}\left[(x^2+1)^\frac{1}{2}+c\right]= \frac{x}{\sqrt{x^2+1}}</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> <math>\frac{da}{dx}\frac{db}{da} = 2x\frac{1}{2}a^{-1/2} = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>")		Revision as of 16:29, 25 August 2022 (view source) Dvaezazizi@laalliance.org (talk \| contribs) No edit summary Newer edit →
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	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>		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>

	<math>\frac{da}{dx}\frac{db}{da} = 2x\frac{1}{2}a^{-1/2} = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>		<math>\frac{da}{dx}\frac{db}{da} = \left(2x\right)\left(\frac{1}{2}a^{-1/2}\right) = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>