`int 1 / ((x-1)sqrt(x^2-2x)) dx` Find or evaluate the integral by completing the square

Friday, April 8, 2016

$\displaystyle\int\frac{{1}}{{{\left({x}-{1}\right)}\sqrt{{{{x}}^{{2}}-{2}{x}}}}}{\left.{d}{x}\right.}$ Find or evaluate the integral by completing the square

We have to evaluate the integral: $\int \frac{1}{(x-1)\sqrt{x^2-2x}}dx$

We can write the integral as:

$\int \frac{1}{(x-1)\sqrt{x^2-2x}}dx=\int \frac{1}{(x-1)\sqrt{(x-1)^2-1}}dx$

Let $\displaystyle{x}-{1}={t}$

So , $\displaystyle{\left.{d}{x}\right.}={\left.{d}{t}\right.}$

hence we can write,

$\int \frac{1}{(x-1)\sqrt{(x-1)^2-1}}dx=\int \frac{1}{t\sqrt{t^2-1}}dt$

Let $\displaystyle{u}={{t}}^{{2}}$

So, $\displaystyle{d}{u}={2}{t}{\left.{d}{t}\right.}$

implies, $dt=\frac{1}{2t}du$

Therefore we have,

$\int\frac{1}{t\sqrt{t^2-1}}dt=\int \frac{1}{t\sqrt{u-1}}.\frac{du}{2t}$

$=\int \frac{1}{2u\sqrt{u-1}}du$

Now let $v=\sqrt{u-1}$

So, $dv=\frac{1}{2\sqrt{u-1}}du=\frac{1}{2v}du$

Hence we have,

$\int \frac{1}{2u\sqrt{u-1}}du=\int \frac{2vdv}{2(v^2+1)v}$

$=\int \frac{dv}{v^2+1}$

$=tan^{-1}(v)+C$ where C is a constant.

$=tan^{-1}(\sqrt{u-1})+C$

$=tan^{-1}(\sqrt{t^2-1})+C$

$=tan^{-1}(\sqrt{(x-1)^2-1})+C$

$=tan^{-1}(\sqrt{x^2-2x})+C$

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