integration - Integral $int{frac{1}{sqrt{2x^2+x+1}}}dx$

I am trying to solve this integral but I can not figure what I do wrong.

$$I=\int{\frac{1}{\sqrt{(2x^2+x+1)}}}dx$$

Here's how I go about it: I think that maybe it can be solved following the

$$\int{\frac{1}{\sqrt{\color{red}{x}^2+\color{blue}{a}^2}}}dx=\ln\left(x+\sqrt{x^2+a^2}\right)$$
I turn the denominator into a sum of 2 products:
$$2x^2+x+1=\left(x\sqrt{2}+\frac{1}{2\sqrt{2}}\right)^2+\left(\frac{\sqrt{7}}{2\sqrt{2}}\right)^2$$
and "$\color{red}{x}$" from the formula would be "$\left(x\sqrt{2}+\frac{1}{2\sqrt{2}}\right)$" while "$\color{blue}a$" would be "$\left(\frac{\sqrt{7}}{2\sqrt{2}}\right)$ also "$x^2+a^2$" is the denominator "$2x^2+x+1$".

When I plug in these numbers I get the following result:

$$I=\ln\left(\left(x\sqrt{2}+\frac{1}{2\sqrt{2}}\right)+{\sqrt{2x^2+x+1}}\right)$$

I sometimes check my results using an online integral calculator and for this one it shows a different result:

$$\frac{\ln\left(\sqrt{\frac{(4x+1)^2}{7}+1}+\frac{4x+1}{\sqrt{7}}\right)}{\sqrt{2}}$$

I am sorry if the formatting is not quite right, It's the best I can do and it took me about an hour aswell. $\ddot \frown$