anonymous
  • anonymous
How do I complete this integration? ∫dx/(√((x^2 )-4x))
Mathematics
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anonymous
  • anonymous
How do I complete this integration? ∫dx/(√((x^2 )-4x))
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
This is messy to do online. You will have to make a couple of substitutions after completing the square of the radicand. So,\[x^2-4x=x^2-4x+(-4/2)^2-(-4/2)^2=(x-2)^2-2^2\]So\[\sqrt{x^2-4x}=\sqrt{2^2[(\frac{x-2}{2})^2-1}]=2\sqrt{(\frac{x-2}{2})^2-1}\]Make the substitution,\[u=\frac{x-2}{2}\]Then,\[2du=dx\]and the integral becomes,\[\frac{1}{2}\int\limits_{}{}\frac{2du}{\sqrt{u^2-1}}=\int\limits_{}{}\frac{du}{\sqrt{u^2-1}}\]Make another substitution,\[u=\cosh(\theta)\]Then\[du=\sinh(\theta)d{\theta}\]and the integral becomes,
anonymous
  • anonymous
\[\int\limits_{}{}\frac{\sinh(\theta)}{\sqrt{\cosh^2(\theta)-1}}d{\theta}=\int\limits_{}{}\frac{\sinh(\theta)}{\sinh(\theta)}d{\theta}=\theta + c\]Then sub. everything back in...\[\theta+c=\cosh^{-1}(u)+c=\cosh^{-1}(\frac{x-2}{2})+c\]
anonymous
  • anonymous
You can use the fact that \[\cosh^{-1}(y)=\ln(y+\sqrt{y^2-1})\]and sub. \[y \rightarrow \frac{x-2}{2}\] and simplify to get something more 'standard'.

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