Finding $\int \frac {dx}{\sqrt {x^2 + 16}}$ I can not get the correct answer. 
$$\int \frac {dx}{\sqrt {x^2 + 16}}$$
$x = 4 \tan \theta$, $dx = 4\sec^2 \theta$
$$\int \frac {dx}{\sqrt {16 \sec^2 \theta}}$$
$$\int \frac {4 \sec^ 2 \theta}{\sqrt {16 \sec^2 \theta}}$$
$$\int \frac {4 \sec^ 2 \theta}{4 \sec \theta}$$
$$\int \sec \theta$$
$$\ln| \sec \theta + \tan \theta|$$
Then I solve for $\theta$:
$x = 4 \tan \theta$
$x/4 = \tan \theta$
$\arctan (\frac{x}{4}) = \theta$
$$\ln| \sec (\arctan (\tfrac{x}{4})) + \tan (\arctan (\tfrac{x}{4}))|$$
$$\ln| \sec (\arctan (\tfrac{x}{4})) + \tfrac{x}{4}))| + c$$
This is wrong and I do not know why.

Cosmetically nicer taking $$ x = 4 \sinh t \; ,  $$
so that $$ dx = 4 \cosh t \; dt $$
The quadratic formula is enought ot give us
$$ t = \log \left( \frac{x + \sqrt{x^2 + 16}}{4} \right) =  \log \left( x + \sqrt{x^2 + 16} \right) - \log 4 $$
Then we get
$$ \int \frac{dx}{x + \sqrt{x^2 + 16} } = \int 1 dt = t + C = \log \left( x + \sqrt{x^2 + 16} \right) - \log 4  + C \; , $$ or
$$ \int \frac{dx}{x + \sqrt{x^2 + 16} } =  \log \left( x + \sqrt{x^2 + 16} \right)  + C_2 \; . $$
