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

Question

Evaluate the given integral

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

Could some give me slight hint as how to initiate this problem?

Answer 1

HINT:

As $\dfrac{x-1}{x^3}=\dfrac1{x^2}-\dfrac1{x^3}$

and $\displaystyle\int\dfrac{x-1}{x^3}dx=-\dfrac1x+\dfrac1{2x^2}+K$

and $2x^2-2x+1=x^2\left(2-\dfrac2x+\dfrac1{x^2}\right)$

Choose $2-\dfrac2x+\dfrac1{x^2}=u$

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Alternatively, choose $x=\dfrac1y$

Answer 2

I would suggest to complete the square under the root and then use a trig substitution.

This is a dirty way to do it but it works most of the time:

$ \begin{align} \int \frac{x-1}{x^2\sqrt{2x^2-2x+1}} \text dx &= \sqrt{2} \int \frac{x-1}{x^2\sqrt{(2x-1)^2+1}} \text dx \\ &= \sqrt{2}\int \frac{\frac{1}{2}(\tan \theta - 1)\sec^2\theta}{(\frac{1}{2})^2(\tan\theta+1)^2\sqrt{\tan^2\theta+1}}\text d\theta\\ &= 2\sqrt{2}\int \frac{(\tan \theta - 1)\sec\theta}{(\tan\theta+1)^2}\text d\theta\\ &= 2\sqrt{2} \int -\frac{\cos\theta-\sin\theta}{(\sin\theta+\cos\theta)^2}\text d\theta\\ &= 2\sqrt{2} \int -\frac{1}{u^2}\text du \end{align} $

I'm fairly certain I made a mistake or two in this but the principle doesn't change. Hope this helps.