`int 1 / ((x-1)sqrt(x^2-2x)) dx` 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 `x-1=t`

So , `dx=dt`

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 `u=t^2`

So, `du=2tdt`

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`