Evaluate $\int \frac{dx}{x\sqrt{4x^2+1}}$ I was evaluating $\int \frac{dx}{x\sqrt{4x^2+1}}$ using Table of Integrals.
My work
I was evaluating $\int \frac{dx}{x\sqrt{4x^2+1}}$ using Table of Integrals. I found in the Table of Integrals an integral that is akin to 
$\int \frac{dx}{x\sqrt{4x^2+1}}$. The integral I was talking about is $$\int \frac{du}{u\sqrt{u^2+a^2}} = -\frac{1}{a} ln \left( \frac{a+\sqrt{u^2+a^2}}{u} \right)$$
With that in mind, I need to modify $\int \frac{dx}{x\sqrt{4x^2+1}}$ to look like $\int \frac{du}{u\sqrt{u^2+a^2}}$.
Modifying now:
$$\int \frac{dx}{x\sqrt{4x^2+1}} = \int \frac{2dx}{2x\sqrt{4x^2+1}} = \int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}}$$
The modified integral to be evaluated would be $\int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}}$. 
Now getting the integral:
$$\int \frac{dx}{x\sqrt{4x^2+1}} = \int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}} = -2\left(\left(\frac{1}{(1)}\right) ln \left( \frac{(1)+\sqrt{(2x)^2+(1)^2}}{(2x)} \right) \right)$$
$$ \int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}} = -2 ln \left( \frac{1+\sqrt{4x^2+1}}{2x} \right) $$
$$ \int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}} =  ln \left(\left( \frac{1+\sqrt{4x^2+1}}{2x} \right)^{-2} \right)$$
$$ \int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}} =  \frac{1}{ln \left(\left( \frac{1+\sqrt{4x^2+1}}{2x} \right)^2 \right)}$$
$$ \int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}} = ln \left( \frac{4x^2}{1+\sqrt{4x^2+1}}\right)$$
So...the integral of $\frac{dx}{x\sqrt{4x^2+1}}$ would be $ln \left( \frac{4x^2}{1+\sqrt{4x^2+1}}\right)$. But in the book I used, it is 
$ln \left( \frac{x}{1+\sqrt{4x^2+1}}\right)$. It was so close to my answer, yet I don't know where I messed up. 
Where did I messed up?

After "Now getting the integral", when you use the formula, you introduce a $2$ that should not be there. You would get
$$
-\ln\left(\frac{1+\sqrt{4x^2+1}}{x}\right)
=\ln\left(\frac{x}{1+\sqrt{4x^2+1}}\right).
$$
You also need to review the rules for logarithms. It is not true that $\ln 1/x=1/\ln x$ (you used it twice, so it actually "cancelled itself", but it is stil wrong). 
