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I try to solve this integral, but without success. Can you help me please?

$$\int \frac{1}{2x+\sqrt{4x^{2}-x+1}}\,dx$$

Thanks a lot!

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    http://www.wolframalpha.com/input/?i=integrate+1%2F%282x%2Bsqrt%284x^2-x%2B1%29%292012-01-31
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    Thanks, but i need also the way of solution, not only the result2012-01-31
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    click on "show steps" (but be prepared...)2012-01-31
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    Sorry, clicking on the above link doesn't quite work; you'll have to copy and paste. How can I get a long link to be clickable in a comment?2012-01-31
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    I'm trying to solve this one... $$\int \frac{2x-\sqrt{4x^{2}-x+1}}{x-1}\;dx$$2012-01-31
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    It's impressive that WA's "show steps" can give such a good simulation of a human mathematician writing out the steps, complete with words explaining what is done in each step. However, this is clearly a case where the machine-generated steps are much too complex for a human to carry out in a reasonable amount of time. I suspect that a human could generate a solution of a more reasonable length. A similar example is $\int dx/(x^{10000}-1)$, which software chokes on but that a human could at least generate some insight into.2012-01-31
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    You can complete the square, make the standard trig substitution, and then, if the trig integral is not easy $t= \tan(\frac{\theta}{2})$ makes it a rational fraction....2012-01-31
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    @David: You can use mini-Markdown formatting as explained when you click on "help" next to the comment text field: [ link text ] ( link url ) (without the spaces) gives [this link](http://www.wolframalpha.com/input/?i=integrate+1%2F%282x%2Bsqrt%284x%5E2-x%2B1%29%29).2012-01-31
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    @Lilly: Since you want to do it using Euler substitution, please tell us what you've tried. E.g., Euler substitution has several different cases. Have you thought about which case or cases might be relevant here?2012-01-31

2 Answers 2

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Hyperbolic substitution
Note $4 x^2 - x +1 = 4\left(x - \frac{1}{8} \right)^2 + \frac{15}{16}$. Therefore, let's perform a $u$-substitution, $x = \frac{1}{8} + \frac{\sqrt{15}}{8} \sinh(t)$. Then $$ 4 x^2 - x +1 = \frac{15}{16} \cosh^2(t) $$ and $\mathrm{d} x = \frac{\sqrt{15}}{8} \cosh(t) \mathrm{d} t$, therefore $$ \int \frac{\mathrm{d} x}{2x + \sqrt{ 4x^2-x+1}} = \int \frac{\frac{\sqrt{15}}{8} \cosh(t)}{\frac{1}{4} + \frac{\sqrt{15}}{4} \sinh(t) + \frac{\sqrt{15}}{4} \cosh(t)} \mathrm{d} t = \frac{\sqrt{15}}{4} \int \frac{ \left( \mathrm{e}^t + \mathrm{e}^{-t}\right)\mathrm{d} t} {1 + \sqrt{15} \mathrm{e}^t} $$ The latter integral is easy $$ \begin{eqnarray} \sqrt{15} \int \frac{\mathrm{e}^t + \mathrm{e}^{-t}}{1+\sqrt{15} \mathrm{e}^t} \mathrm{d} t &=& \int \left( \frac{\sqrt{15}}{\mathrm{e}^{t} } - 15 + \frac{16 \sqrt{15} \mathrm{e}^t }{1 + \sqrt{15} \mathrm{e}^t} \right) \mathrm{d} t \\ &=& -\sqrt{15} \mathrm{e}^{-t} - 15 t + 16 \log\left( 1 + \sqrt{15} \mathrm{e}^t\right) + \color{\gray}{\text{const}} \end{eqnarray} $$ Back-substitution of $t = \sinh^{-1}\left(\frac{8x-1}{\sqrt{15}} \right)$, and using $\mathrm{e}^{-\sinh^{-1}(z)} = \sqrt{1+z^2}-z$ gives $$ \int \frac{\mathrm{d} x}{2x + \sqrt{ 4x^2-x+1}} = \frac{1}{4} (8 x-1)-\frac{1}{4} \sqrt{(8 x-1)^2 + 15}+\frac{1}{4} \sinh ^{-1}\left(\frac{8 x-1}{\sqrt{15}}\right)+4 \log \left(15-(8 x-1)+\sqrt{(8 x-1)^2+15}\right) + \color{\gray}{\text{const}} $$

Euler substitution
Let $u = 2x + \sqrt{4 x^2 - x+1}$. Notice that $(u-2x)^2 = u^2 - 4 x u + 4 x^2 = 4 x^2 - x + 1$, that makes it $$ x = \frac{u^2-1}{4 u-1} = \frac{1}{16} + \frac{u}{4} - \frac{15}{16(4u-1)} $$ Therefore $$ \mathrm{d} x = \frac{\mathrm{d} u}{4} + \frac{15}{4} \frac{\mathrm{d} u}{(4u-1)^2} $$ Using the above $$ \begin{eqnarray} \int\frac{\mathrm{d} x}{2x + \sqrt{4x^2-x+1}} &=& \int \frac{1}{u} \left( \frac{1}{4} + \frac{15}{4} \frac{1}{(4u-1)^2} \right) \mathrm{d} u \\ &=& \int \left( \frac{4}{u} + \frac{15}{(4u-1)^2} - \frac{15}{4u-1} \right) \mathrm{d} u \\ &=& 4 \ln(u) - \frac{15}{4(4u-1)} - \frac{15}{4} \ln(1-4u) + \color{\gray}{\text{const}} \end{eqnarray} $$

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    Sweet job on this, but the OP specifically wants to do it using Euler substitution, not a hyperbolic substitution.2012-01-31
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    +1, You might as well reply to this prior question concerning the same integral ["Integral without using Euler substitution"](http://math.stackexchange.com/questions/103227/integral-without-using-euler-substitution). As far as I understand the Euler substitution, e.g. entry [Euler substitutions](http://www.encyclopediaofmath.org/index.php/Euler_substitutions) of Springer Encyclopedia of Mathematics for $a>0$ is $\sqrt{ax^2+bx+c}=t-\sqrt{a}x$.2012-01-31
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    @AméricoTavares My bad, I was not paying attention. I have added the Euler substitution solution as well.2012-01-31
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    Very nice indeed.2012-01-31
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    @Sasha Thanks a lot!2012-02-01
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Go to the link

http://www.wolframalpha.com/input/?i=1%2F%282x+%2B+sqrt%284x%5E2-x%2B1%29%29

and you will find a lot of facts about your function, including the integral you're looking for.

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    OK, now I see someone else had already tried Wolfram Alpha. Sorry.2012-01-31