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anonymous
 5 years ago
How do I complete this integration?
∫dx/(√((x^2 )4x))
anonymous
 5 years ago
How do I complete this integration? ∫dx/(√((x^2 )4x))

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is messy to do online. You will have to make a couple of substitutions after completing the square of the radicand. So,\[x^24x=x^24x+(4/2)^2(4/2)^2=(x2)^22^2\]So\[\sqrt{x^24x}=\sqrt{2^2[(\frac{x2}{2})^21}]=2\sqrt{(\frac{x2}{2})^21}\]Make the substitution,\[u=\frac{x2}{2}\]Then,\[2du=dx\]and the integral becomes,\[\frac{1}{2}\int\limits_{}{}\frac{2du}{\sqrt{u^21}}=\int\limits_{}{}\frac{du}{\sqrt{u^21}}\]Make another substitution,\[u=\cosh(\theta)\]Then\[du=\sinh(\theta)d{\theta}\]and the integral becomes,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}{}\frac{\sinh(\theta)}{\sqrt{\cosh^2(\theta)1}}d{\theta}=\int\limits_{}{}\frac{\sinh(\theta)}{\sinh(\theta)}d{\theta}=\theta + c\]Then sub. everything back in...\[\theta+c=\cosh^{1}(u)+c=\cosh^{1}(\frac{x2}{2})+c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can use the fact that \[\cosh^{1}(y)=\ln(y+\sqrt{y^21})\]and sub. \[y \rightarrow \frac{x2}{2}\] and simplify to get something more 'standard'.
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