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.
 A: What you have done is correct! Note that whenever you have inverse trigonometric expressions you can express your answer in more than one way! Your answer can be expressed in a different way (without the trigonometric and inverse trigonometric functions) as shown below.
We will prove that $$\sec \left( \arctan \left( \dfrac{x}4 \right) \right) = \sqrt{1 + \left(\dfrac{x}{4} \right)^2}$$
Hence, your answer $$\ln \left \lvert \dfrac{x}4 + \sec \left(\arctan \left( \dfrac{x} 4\right) \right) \right \rvert + c$$ can be rewritten as $$\ln \left \lvert \dfrac{x}4 + \sqrt{1+ \left(\dfrac{x}{4} \right)^2} \right \rvert + c$$
Note that $$\theta = \arctan\left( \dfrac{x}4 \right) \implies \tan( \theta) = \dfrac{x}4 \implies \tan^2(\theta) = \dfrac{x^2}{16} \implies 1 + \tan^2(\theta) = 1+\dfrac{x^2}{16}$$
Hence, we get that $$\sec^2(\theta) = 1+ \left(\dfrac{x}{4} \right)^2 \implies \sec (\theta) = \sqrt{1+ \left(\dfrac{x}{4} \right)^2} \implies \sec \left(\arctan\left( \dfrac{x}4 \right) \right) = \sqrt{1+ \left(\dfrac{x}{4} \right)^2}$$
Hence, you can rewrite your answer as
$$\ln \left \lvert \dfrac{x}4 + \sqrt{1+ \left(\dfrac{x}{4} \right)^2} \right \rvert + c$$
Also, you have been a bit sloppy with some notations in your argument.
For instance, when you substitute $x = 4 \tan (\theta)$,
$$\dfrac{dx}{\sqrt{x^2+16}} \text{ should immediately become }\dfrac{4 \sec^2(\theta)}{\sqrt{16 \sec^2(\theta)}} d \theta$$
Also, you need to carry the $d \theta$ throughout the answer under the integral.
Writing just $\displaystyle \int\sec(\theta)$ or $\displaystyle \int\dfrac{4 \sec^2(\theta)}{\sqrt{16 \sec^2(\theta)}}$ without the $d \theta$ is notationally incorrect.
Anyway, I am happy that you are slowly getting a hang of these!
A: 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
$$ \int \frac{dx}{x + \sqrt{x^2 + 16} } =  \log \left( x + \sqrt{x^2 + 16} \right)  + C_2 \; . $$
A: $$\int \frac {dx}{\sqrt {x^2 + 16}}$$
Let $$t=\sqrt {x^2+16}$$ this will change the integral into $$ \int \frac {dt}{\sqrt {t^2 - 16}}$$
$$t=4\sec (\theta)$$ changes the integral into $$\int \sec \theta d \theta =ln| \sec \theta + \tan \theta|+c$$
$$=\ln (t/4 + \frac {\sqrt {t^2-16}}{4} )+c$$
$$= \ln ( \sqrt \frac {x^2+16}{4}+x/4 )+c$$
