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 step is put $x = u+\dfrac{1}{2}\Rightarrow \sqrt{x^2+(x-1)^2}=\sqrt{\left(u+\dfrac{1}{2}\right)^2+\left(u-\dfrac{1}{2}\right)^2}=\sqrt{2u^2+\dfrac{1}{2}}=\sqrt{2}\cdot \sqrt{u^2+\dfrac{1}{4}}$. Second step is to put $u =\dfrac{1}{2}\tan \theta$. I think we can go further at this point.
