Integrate $\int{ \frac{x-1}{x^2\sqrt{2x^2-2x+1}} dx}$

Question

$$\int{ \frac{x-1}{x^2\sqrt{2x^2-2x+1}} \, \text{d} x}$$ Is there a clever way to integrate this? I know the answer is $\frac{\sqrt{2x^2-2x+1}}{x}$, but I have no idea how to calculate it myself.

I have transformed it into $\sqrt2 \int{ \frac{x-1}{x^2\sqrt{(2x-1)^2+1}} \, \text{d} x}$. Then I thought about splitting the integral into $\int{ \frac{1}{x \sqrt{2x^2-2x+1}} \, \text{d} x}$ and $\int{ \frac{-1}{x^2\sqrt{2x^2-2x+1}} \, \text{d} x}$, but they are not much easier for me. Tried substituting $\tan x$ for $2x-1$ into the first one, but arrived at $\int{ \frac{\cos t}{\sin{2t} + 1} \, \text{d} t}$ and I don't like it.

The simple solution makes me think that there should be a clever substitution or some way to transform the integrand so that it can be solved quickly. I would appreciate some hint, as well as any general tips on solving this kind of integrals. I know it can be done with Euler's substitutions, but I tend to make mistakes in there and get wrong answers after 2 pages of calculations, which is not very encouraging.

Answer 1

We can compute as follows. $$\begin{split} &\ \int{ \frac{x-1}{x^2\sqrt{2x^2-2x+1}} \, dx}\\ &=\int{ \frac{1-x^{-1}}{x^2\sqrt{2-2x^{-1}+x^{-2}}} \, dx}\\ &=\int{ \left(\frac{1}{x^2}-\frac{1}{x^3}\right) \frac{1}{\sqrt{2-2x^{-1}+x^{-2}}} \, dx}\\ &=\int{ \frac{1}{2}\frac{dt}{dx}\frac{1}{\sqrt{t}} \, dx}\\ &=\sqrt{t}+C \\ &=\sqrt{2-2x^{-1}+x^{-2}} \end{split}$$ where $t=2-2x^{-1}+x^{-2}$.

Answer 2

Substitute $y = \sqrt{2x^2-2x+1}$ to have $ ydy=(2x-1)dx$

\begin{align} &\int{ \frac{x-1}{x^2\sqrt{2x^2-2x+1}} dx}\\ =& \int \frac{x(2x-1)-y^2}{x^2y} dx =\int \frac{xdy-ydx}{x^2}=\int d\left(\frac yx\right)=\frac yx \end{align}