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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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  • $\begingroup$ wolframalpha.com/input/?i=integrate+1%2F%282x%2Bsqrt%284x^2-x%2B1%29%29 $\endgroup$ – David Mitra Jan 31 '12 at 15:54
  • $\begingroup$ Thanks, but i need also the way of solution, not only the result $\endgroup$ – Lilly Jan 31 '12 at 16:07
  • $\begingroup$ click on "show steps" (but be prepared...) $\endgroup$ – David Mitra Jan 31 '12 at 16:08
  • $\begingroup$ 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? $\endgroup$ – David Mitra Jan 31 '12 at 16:12
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    $\begingroup$ @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? $\endgroup$ – Ben Crowell Jan 31 '12 at 18:20
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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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  • $\begingroup$ Sweet job on this, but the OP specifically wants to do it using Euler substitution, not a hyperbolic substitution. $\endgroup$ – Ben Crowell Jan 31 '12 at 18:17
  • $\begingroup$ +1, You might as well reply to this prior question concerning the same integral "Integral without using Euler substitution". As far as I understand the Euler substitution, e.g. entry Euler substitutions of Springer Encyclopedia of Mathematics for $a>0$ is $\sqrt{ax^2+bx+c}=t-\sqrt{a}x$. $\endgroup$ – Américo Tavares Jan 31 '12 at 18:19
  • $\begingroup$ @AméricoTavares My bad, I was not paying attention. I have added the Euler substitution solution as well. $\endgroup$ – Sasha Jan 31 '12 at 18:36
  • $\begingroup$ Very nice indeed. $\endgroup$ – Américo Tavares Jan 31 '12 at 19:45
  • $\begingroup$ @Sasha Thanks a lot! $\endgroup$ – Lilly Feb 1 '12 at 11:10
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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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  • $\begingroup$ OK, now I see someone else had already tried Wolfram Alpha. Sorry. $\endgroup$ – Manolito Pérez Jan 31 '12 at 17:11

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