Integrating $\int \frac {dx}{\sqrt{4x^{2}+1}}$ $\int \dfrac {dx}{\sqrt{4x^{2}+1}}$
I've been up to this one for quite a while already, and have tried several ways to integrate it, using substituion, with trigonometric as well as hyperbolic functions. I know(I think) I'm supposed to obtain:
$\dfrac {1}{2}\ln \left| 2\sqrt {x^2 +\dfrac {1}{4}}+2x\right|+c$
However, whatever idea I come up with to try to solve it, it seems I always obtain: 
$\dfrac {1}{2}\ln \left| x + \sqrt {x^2 + 1} \right| + c$
Thanks for reading, and hopefully answering too.
Cheers
Yann

we have $\int \dfrac {dx}{\sqrt{4x^{2}+1}}=\int \dfrac {x}{x\sqrt{4x^{2}+1}}\,dx$.Now let $1+4x^{2}=u^2$, so we have $xdx=\frac{1}{4}udu$ and so $$\int \dfrac {x}{x\sqrt{4x^{2}+1}}\,dx=\frac{1}{4}\int \dfrac {u}{u\frac{u^2-1}{4}}\,du=\int \dfrac {1}{u^2-1}\,du$$ also $$\int \dfrac {1}{u^2-1}\,du=\frac{1}{2}\int (\frac{1}{u-1}-\frac{1}{u+1})\,du=\frac{1}{2}\ln|\frac{u-1}{u+1}|.$$ Therefore $$\int \dfrac {dx}{\sqrt{4x^{2}+1}}=\frac{1}{2}\int (\frac{1}{u-1}-\frac{1}{u+1})\,du=\frac{1}{2}\ln|\frac{u-1}{u+1}|+C.$$
