Computing $\int \frac{dx}{(x-1)\sqrt{2x^2-2x+1}}$

Question

I need to compute the following integral: $$\int \frac{dx}{(x-1)\sqrt{2x^2-2x+1}}$$

My try:

I saw that the derivative of $\sqrt{2x^2-2x+1}$ is $\frac{2 x - 1}{\sqrt{2 x^2 - 2 x + 1}}$, so I used $u=\sqrt{2x^2-2x+1}$, $du=\frac{2 x - 1}{\sqrt{2 x^2 - 2 x + 1}} dx$. Then the integral is: $$\int \frac{du}{(x-1)(2x-1)} = \int \frac{du}{2x^2-3x+1} $$

So my try doesn't work (I thought I would get $u^2$ or something like that in the denominator, but I'm off by $x$)

Any hints are appreciated.

Answer 1

As $2x^2-2x+1=\dfrac{(2x-1)^2+1^2}2$

using Trigonometric substitution, $2x-1=\tan t$ where $-\dfrac\pi2\le t\le\dfrac\pi2$

$$\int\dfrac{dx}{(x-1)\sqrt{2x^2-2x+1}}=\sqrt2\int\dfrac{\sec^2t\ dt}{(\tan t-1)\sec t}=\sqrt2\int\dfrac{dt}{\sin t-\cos t} =\int\dfrac{dt}{\sin\left(t-\dfrac\pi4\right)}$$

Answer 2

Hint: Substitute $$\sqrt{2x^2-2x+1}=\sqrt{2}x+t$$ the so called Euler substitution.

Answer 3

Hint: can also take substitution as Let $$(x-1)=\frac{1}{t}$$ Integration will get reduced to form $$\int \frac{dx}{\sqrt{at^2+bt+c}}$$

Answer 4

I would go this way. \begin{eqnarray} \mathcal I &=& \int \frac1{(x-1)\sqrt{2x^2-2x+1}}dx=\\ &=&\frac1{\sqrt 2}\int\frac1{(x-1)\sqrt{\left(x-\frac12\right)^2+\frac14}}dx. \end{eqnarray} Then substitute $x-\frac12 = \frac12 t$, to get \begin{eqnarray} \mathcal I &=& \frac1{2\sqrt 2}\int\frac{1}{\left(\frac{t}2-\frac12\right)\sqrt{\frac{t^2}4+\frac14}}dt=\\ &=&\sqrt 2\int \frac{1}{(t-1)\sqrt{t^2+1}}dt. \end{eqnarray}

Answer 5

Substitute $x=\frac y{1+y}$ $$\int \frac{dx}{(x-1)\sqrt{2x^2-2x+1}} =- \int \frac1{\sqrt{1+y^2}}dy=-\sinh^{-1}y $$