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

1. 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,

2. 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...$\theta+c=\cosh^{-1}(u)+c=\cosh^{-1}(\frac{x-2}{2})+c$

3. 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'.