Find $\int_{1}^{2}\frac{x-1}{x^2\sqrt{x^2+(x-1)^2}}$ Find $\int_{1}^{2}\frac{x-1}{x^2\sqrt{x^2+(x-1)^2}}$

I tried to solve it by using the property $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$
Let $I=\int_{1}^{2}\frac{(x-1)dx}{x^2\sqrt{x^2+(x-1)^2}}$
$I=\int_{1}^{2}\frac{(2-x)dx}{(3-x)^2\sqrt{(3-x)^2+(2-x)^2}}$
But this does not seem to work here.I am stuck.What should i do?

First write $$x - 1 = x(2x - 1) - (2x^2 - 2x + 1).$$  Then our integrand becomes:  $$\frac{x-1}{x^2 \sqrt{2x^2 - 2x + 1}} = \frac{2x-1}{x\sqrt{2x^2 - 2x + 1}} - \frac{\sqrt{2x^2 - 2x + 1}}{x^2}.$$  Now with the choice $$u = \sqrt{2x^2 - 2x + 1}, \quad v = \frac{1}{x},$$ we have $$u' = \frac{2x-1}{\sqrt{2x^2-2x+1}}, \quad v' = -\frac{1}{x^2},$$ hence the integrand is of the form $$u'v + uv' = (uv)'.$$  It immediately follows that an antiderivative is $$\int \frac{x-1}{x^2\sqrt{2x^2-2x+1}} \, dx = uv + C = \frac{\sqrt{2x^2 - 2x + 1}}{x} + C.$$  The rest is straightforward.
