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Help me please with integral: $$\int \frac{2x-\sqrt{4x^{2}-x+1}}{x-1}\;dx$$

I must solve it without using Euler substitution.


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What is Euler substitution? – Gerry Myerson Jan 28 '12 at 11:17
@Gerry Myerson: In this case ($a>0$) the Euler substitution is $\sqrt{ax^2+bx+c}=t-x\sqrt{a}$. See Springer Encyclopedia of Mathematics, Euler substitutions. – Américo Tavares Jan 28 '12 at 14:48
I deleted my wrong answer, because I was not able to find a correct solution. – Américo Tavares Jan 28 '12 at 22:46
No trig substitutions either? – N. S. Jan 28 '12 at 23:11
Is it solvable by Euler substitution? I can't solve it at all... – Lilly Jan 31 '12 at 15:33
up vote 6 down vote accepted

Hint: writing $$\frac{2x-\sqrt{4x^2-x+1}}{x-1}=\frac{2x-\sqrt{4x^2-x+1}}{x-1}\frac{2x+\sqrt{4x^2-x+1}}{2x+\sqrt{4x^2-x+1}},$$ we find $$\frac{2x-\sqrt{4x^2-x+1}}{x-1}=\frac{4x^2-(4x^2-x+1)}{(x-1)(2x+\sqrt{4x^2-x+1})}=\frac 1{2x+\sqrt{4x^2-x+1}}.$$

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I think I see a couple mistakes here. In the first line, if we're rationalizing the numerator, there's a sign error where the denominator should be $2x + \sqrt{4x^2 - x + 1}$, and in the second line the $2x$ term seems to have disappeared from the denominator completely. The end result should be $\frac{1}{2x + \sqrt{4x^2 - x + 1}}$, but I think this ruins the approach expanded upon by @Américo Tavares. – kate Jan 28 '12 at 20:13
@lewist: Yes, it ruins. I am not able to find a new valid solution. – Américo Tavares Jan 28 '12 at 22:21

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