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Prove that $ \Large\frac{d}{dx}\bigg[\Large\frac{1}{4 \sqrt 2} \normalsize log\bigg|\Large\frac{x^2+\sqrt 2 x+1}{x^2- \sqrt 2 x+1} \bigg|+\Large\frac {1}{2 \sqrt 2} \normalsize \tan^{-1} \Large\frac{\sqrt 2 x}{1-x^2}\bigg]=\Large\frac{1}{1+x^4}$

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Toolbox:
  • Write $ log \: \large\frac{a}{b}=log\: a - log\: b$
  • Simplify to get the result
$ \large\frac{d}{dx} \bigg[\large \frac{1}{4\sqrt 2} ( log\: x^2+\sqrt 2x+1-log\: x^2-\sqrt 2x+1 ) + \large\frac{1}{2\sqrt 2}tan^{-1}\large\frac{\sqrt 2x}{1-x^2} \bigg]$
 
$ = \large\frac{1}{4\sqrt 2} \bigg[ \large\frac{2x+\sqrt2}{x^2+\sqrt 2x+1}-\large\frac{2x-\sqrt2}{x^2-\sqrt 2x+1} \bigg] + \large\frac{1}{2\sqrt 2}\large \frac{(1-x^2)^2}{1+x^4}$x$\large\frac{(1-x^2)\sqrt 2+ 2\sqrt 2x^2}{(1-x^2)^2}$
 
Simplify and get $ \large\frac{1}{1+x^4}$

 

answered Mar 9, 2013 by thanvigandhi_1
edited Mar 25, 2013 by thanvigandhi_1
 

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