Calculus of a Single Variable, Chapter 5, 5.7, Section 5.7, Problem 40

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