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

Latest revision as of 18:24, 26 August 2022

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

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

${\begin{aligned}a&=x^{2}+1&b&=a^{1/2}\\[0.6ex]{\frac {da}{dx}}&=2x&{\frac {db}{da}}&={\frac {1}{2}}a^{-1/2}\end{aligned}}$

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

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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>
-<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>
-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>
+<br>
+<math>\begin{align}
+a &= x^2+1 & b &= a^{1/2} \\[0.6ex]
+\frac{da}{dx} &=2x & \frac{db}{da} &=\frac{1}{2}a^{-1/2}
+\end{align}</math>
+<br><br>
-<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}\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>