My Math Forum Trying to understand integration with very basic knowledge of derivation

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March 6th, 2018, 08:25 AM   #1
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Trying to understand integration with very basic knowledge of derivation

I have to show that this equation is true, I know I have to find the derivative of the right side, but I don't know how to find derivative of the square root.
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 March 6th, 2018, 11:50 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Do you know the power rule? The derivative of $x^n$ is $nx^{n-1}$. $\sqrt{x}= x^{1/2}$ so the derivative is $\frac{1}{2}x^{-1/2}= \frac{1}{2\sqrt{x}}$. Further, the derivative of ln(x) is $\frac{1}{x}$. Using the "chain rule", the derivative of $ln\left(x+ \sqrt{x^2+ 1}\right)$ is $\frac{1}{x+ \sqrt{x^2+ 1}}$ times the derivative of $x+ \sqrt{x^2+ 1}= x+ (x^2+ 1)^{1/2}$. And that derivative is $1+ (1/2)(x^2+1)^{-1/2}(2x)$$= 1+ \frac{x}{\sqrt{x^2+ 1}}. So the derivative you seek is \frac{1+ \frac{x}{\sqrt{x^2+ 1}}}{x+ \sqrt{x^2+ 1}} Multiplying numerator and denominator by \sqrt{x^2+ 1} simplifies it slightly to \frac{x+ \sqrt{x^2+ 1}}{x\sqrt{x^2+ 1}+ 1} Thanks from greg1313 and Sebastian Garth Last edited by greg1313; March 6th, 2018 at 12:36 PM.  March 6th, 2018, 12:41 PM #3 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,950 Thanks: 1141 Math Focus: Elementary mathematics and beyond$$\frac{1+ \frac{x}{\sqrt{x^2+ 1}}}{x+ \sqrt{x^2+ 1}}=\frac{\frac{\sqrt{x^2+1}+x}{ \sqrt{x^2+1}}}{x+\sqrt{x^2+1}}=\frac{1}{\sqrt{x^2+ 1}}$\$

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