<s>[INST] Integral( using Euler substitution) 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!
 [/INST] 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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