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I've just tried to use the Euler's formula for my integral, but I can't get the correct answer. So if anyone could help me I would really appreciate that. This is my integral: $$\int\frac{\sqrt{x^2+2x-1} }{x}\,dx$$

P.S. The ingral must be solven using Euler's formula

This is where I've got stuck: I started with this substitution: $\sqrt{x^2+2x-1} = -x + t$. After derivating I get $dx= t^2 + 2x -1 /2(t+1)^2$. After immpleneting it into my integral I get to this point $$\int\frac{(t^2+2t-1)(t^2+2t-1)}{(t^2+1)2(t+1)^2}\,dt$$ and I don't have any idea what I should do next (thought to do another substitution but don't know what to substitute).

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  • $\begingroup$ Well..I started with this substitution: $\sqrt{x^2+2x-1}$ = -x + t . After derivating I get dx= t^2 + 2x -1 /2(t+1)^2 After immpleneting it into my integral I get to this point $\int_{}^{}\frac{(t^2+2t-1)*(t^2+2t-1)}{(t^2+1)2(t+1)^2}$ and I don't have any idea what I should do next (thought to do another substitution but don't know what to substitute) @Moo $\endgroup$ – Dragan Zrilić Mar 14 '16 at 17:32
  • $\begingroup$ @DraganZrilić Since this quadratic has two distinct real roots, Euler's third substitution will probably offer the cleanest solution. $\endgroup$ – David H Mar 14 '16 at 18:23
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    $\begingroup$ F**k Euler's substitution. It's a mess. :) $\endgroup$ – Von Neumann Mar 14 '16 at 18:49
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After we get $$\int\frac{(t^2+2t-1)(t^2+2t-1)}{(t^2+1)2(t+1)^2}\,dt$$

We can try to transform it into alternate form. Just let

$$\frac{(t^2+2t-1)(t^2+2t-1)}{(t^2+1)2(t+1)^2} = a + \frac{b}{t+1} + \frac{ct+d}{(t+1)^2} + \frac{e}{t^2+1}$$

We can reduct the right to a common denominator.Then compare the coefficient.

Or we can substitute some t then we can get a lot of linear equation then solve them.(For convinience, we can substitute $t=0$,$t=1$,$t=2$,$t=3$,$t=4$,then get five equations.)

After solved, we get: $$ \left\{ \begin{array}{c} a=\frac{1}{2}\\ b=1\\ c=0\\ d=1\\ e=-2 \end{array} \right.$$

Then it's tranformed into (I do it by WA)

$$\int\left(\frac{1}{2} + \frac{1}{t+1} + \frac{1}{(t+1)^2} - \frac{2}{t^2+1}\right)dt$$

And now we can get the answer:

$$\frac{t}{2} + ln(t+1) - \frac{1}{t+1} - 2arctan(t)$$

It's hard and boring to caculate. But it works(●'◡'●)

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