Problem integrating (substitution)

Question

can you help me identify the mistake I'm making while integrating?

Question:

$$\int{\frac{2dx}{x\sqrt{4x^2-1}}}, x>\frac{1}{2}$$

my solution

$$\int{\frac{2dx}{x\sqrt{4x^2-1}}}=2\int{\frac{dx}{x\sqrt{(2x)^2-1}}}$$

let $$u=2x, x=1/2u, du=2dx, 1/2du=dx$$

$$=\frac{2}{2}\int{\frac{du}{1/2u\sqrt{u^2-1}}}$$

$$=2\int{\frac{du}{u\sqrt{u^2-1}}}$$

It is known $\int{\frac{dx}{x\sqrt{x^2-a^{2}}}}=\frac{1}{a}sec^{-1}{|\frac{x}{a}|}+C$

so

$$=2\int{\frac{du}{u\sqrt{u^2-1}}}=2(sec^{-1}u)+C$$

$$=2(sec^{-1}2x)+C$$

unfortunately Wolfram Alpha says the answer is $$-2(tan^{-1}\frac{1}{\sqrt{4x^{2}-1}})+C$$

1. Are these answers equivalent?

2. What identities should i use to test equivalence?

3. If i made a mistake, where is it?

Thanks staxers

Answer 1

Remember the identity: $$\sec^2 \alpha=1+\tan^2\alpha$$

Using this, it is not difficult to show that both expressions are equivalent.

EDIT:

Let's say that $u=\sec^{-1}2x$. Then $2x=\sec u=\sqrt{1+\tan^2u}$. Thus, $$4x^2=1+\tan^2u$$ $$\tan^2u=4x^2-1$$ $$\tan u=\sqrt{4x^2-1}$$ $$u=\tan^{-1}\sqrt{4x^2-1}$$

Answer 2

$$\begin{align} \int{\frac2{\color{red}x\sqrt{4x^2-1}}}~dx ~&=~ \int{\frac{2\color{red}x}{\color{red}{x^2}\sqrt{\color{blue}{4x^2-1}}}}~dx ~=~ \int\frac{\dfrac14\cdot du}{\bigg(\dfrac{u+1}4\bigg)\cdot\sqrt{\color{blue}u}} ~=~ \int\frac{du}{(u+1)\cdot\sqrt u} ~=~ \\\\ ~&=~ \int\frac{2t}{(t^2+1)~t}~dt ~=~ 2\int\frac{dt}{t^2+1} ~=~ 2\arctan t. \end{align}$$