Evaluate the integrals in terms of inverse hyperbolic functions & natural logs Having trouble with my homework. It is asking evaluate the integral in terms of 1) inverse hyperbolic function and 2) natural logarithm
The integral is $$ \int_0^{2\sqrt3} \frac{dx}{\sqrt{4+x^2}} $$
I would also post what I have so far, but I don't have anything :/ Any help would be very very appreciated. 
 A: $$
\sqrt{4+x^2} = \sqrt 4 \cdot \sqrt{1+ \left( \frac x 2 \right)^2}.
$$
When you see "$1+(\cdots)^2$", you think of a certain trigonometric substitution.
(And when you see "$1-(\cdots)^2$", you think of a certain other trigonometric substitution.)
(And when you see "$(\cdots)^2 -1$", you think of a certain other trigonometric substitution.)
PS: Alright, let's pursue this:
\begin{align}
\frac x 2 & = \tan\theta \\[10pt]
\frac{dx} 2 & = \sec^2\theta\,d\theta \\[10pt]
1 + \tan^2\theta & = \sec^2\theta
\end{align}
When $\theta=0$ then $x=0$, and when $\theta=\pi/3$ then $x=2 \sqrt 3$. So
\begin{align}
\int_0^{2\sqrt 3} \frac{dx/2}{\sqrt{1 + (x/2)^2}} & = \int_0^{\pi/3} \frac{\sec^2\theta\,d\theta}{\sec\theta} = \int_0^{\pi/3} \sec\theta\,d\theta \\[10pt]
& = \left. \log(\sec\theta+\tan\theta) \vphantom{\frac 1 1} \right|_0^{\pi/3} = \log(2+\sqrt 3).
\end{align}
Is this the same as $\sinh^{-1}\sqrt3\,{}$?  Let's check:
$$
\sinh\log(2+\sqrt3) = \frac{e^{\log(2+\sqrt 3)} + e^{-(\log(2+\sqrt 3))}} 2 = \frac {2+\sqrt 3 - \frac 1 {2+\sqrt 3}} 2.
$$
By rationalizing the denominator we find that this is the same as
$$
\frac {(2+\sqrt 3) - (2-\sqrt 3)} 2 = \sqrt 3.
$$
So it checks.
A: method 1: $$u = \sqrt{4+x^2} \to u^2 = 4+x^2 \to x^2 = u^2-4\to x = \sqrt{u^2-4}\to 2u\,du = 2x\,dx\to u\,du = x\,dx\to I = \int_2^4 \dfrac{du}{\sqrt{u^2-4}}.$$ From here we can put $u = 2\cosh(v)$
method 2: 
put $x = 2\tan(u)$
method 3:
put $x = 2\sinh(u)$
A: Hint: Substitute $x=2\cosh \theta$
