5.4 Indefinite Integrals and the Net Change Theorem/1: Difference between revisions
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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><br> | ||
Show that: <math>\frac{d}{dx}\left[(x^2+1)^\frac{1}{2}+c\right]= \frac{x}{\sqrt{x^2+1}}</math> | Show that: <math>\frac{d}{dx}\left[(x^2+1)^\frac{1}{2}+c\right]= \frac{x}{\sqrt{x^2+1}}</math> | ||
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<math>\begin{align} | <math>\begin{align} | ||
a &= x^2+1 & b &= a^{1/2} \\[0.6ex] | a &= x^2+1 & b &= a^{1/2} \\[0.6ex] | ||
\frac{da}{dx} &=2x | \frac{da}{dx} &=2x & \frac{db}{da} &=\frac{1}{2}a^{-1/2} | ||
\end{align}</math> | \end{align}</math> | ||
<br><br> | <br><br> | ||
<math>\frac{da}{dx}\cdot\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> | <math>\frac{da}{dx}\cdot\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> | ||
Latest revision as of 18:24, 26 August 2022
Show that:
![{\displaystyle {\frac {d}{dx}}\left[(x^{2}+1)^{\frac {1}{2}}+c\right]={\frac {x}{\sqrt {x^{2}+1}}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/3554e3a809bdd72518b30ec0c10b7cb78eba7739)
![{\displaystyle {\begin{aligned}a&=x^{2}+1&b&=a^{1/2}\\[0.6ex]{\frac {da}{dx}}&=2x&{\frac {db}{da}}&={\frac {1}{2}}a^{-1/2}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/a3c658000d63c4d460517b1c89cddd4523016562)
