Compute $\int{\frac{dx}{\sqrt{x^2+16}}}$ Here is what I have done so far:
$$I=\int \frac{dx}{\sqrt{x^2+16}} =\frac{1}{4}\int\sqrt{\frac{1}{\left(\frac{x}{4} \right)^2+1}}\,dx$$
$\frac{x}{4}=\tan(u), dx=4\sec^2(u)$
$$\therefore I=\int\sqrt{\frac{1}{\tan^2(u)+1}}\sec^2(u)\,du=\int\sec^3(u)\,du$$
now by integration by parts:
$$I=\tan(u)\sec(u)-\int\tan^2(u)\sec(u)\,du$$
I am sure $\int\sec^3(u)du$ is a standard integral but I am now sure if IBP is the best way to go. Is there another obvious way of doing it?
 A: Hint: Substitute
$$\sqrt{x^2+16}=xt+4$$ this is the so-called Euler substitution.
See here https://en.wikipedia.org/wiki/Euler_substitution
A: *

*Either you've learnt about   the inverse hyperbolic functions, and you know that
$$\int\frac{\mathrm d x}{\sqrt{x^2+1}}=\arg\sinh x=\ln\bigl(x+\sqrt{x^2+1}\bigr),$$
from which one deduces readily with the substitution  $x=at$ ($a>0$) that
$$\int\frac{\mathrm d x}{\sqrt{x^2+a^2}}=\arg\sinh \frac xa=\ln\bigl(x+\sqrt{x^2+a^2}\bigr).$$

*Or you've only learnt about the hyperbolic functions. In this case, set $x=a\sinh t$, $\mathrm d x=a\cosh t\,\mathrm d t$, so
$$\int\frac{\mathrm d x}{\sqrt{x^2+a^2}}=\int\frac{a\cosh t\,\mathrm d t}{a\sqrt{\sinh^2 t+1}}=\int\frac{\cosh t\,\mathrm d t}{\cosh t}=t.$$
There remains to solve for $x$ the equation
$$t=\sinh x=\frac{\mathrm e^{2x}-1}{2\mathrm e^x}.$$

A: Use the trigonometric substitution as follows:
Let $$x = 4\times \tan(\theta)$$
By doing the necessary operations you will end up with $$\int \sec (\theta)\, d\theta $$
Which i guess you can easily solve  
