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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!

  • 1
    @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} $

  • 0
    @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.

  • 0
    OK, now I see someone else had already tried Wolfram Alpha. Sorry.2012-01-31