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

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-<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>

Latest revision as of 19:59, 23 August 2022