Evaluate the Integral: $\int\frac{dx}{\sqrt{x^2+16}}$ I want to evaluate $\int\frac{dx}{\sqrt{x^2+16}}$.
My answer is: $\ln \left| \frac{4+x}{4}+\frac{x}{4} \right|+C$
My work is attached: 

$u = \ln \left| {x + \sqrt {{x^2} + 16} } \right|$. Then $du = \frac{1}
{{\sqrt {{x^2} + 16} }}dx$. Then
$$\int {\frac{1}{{\sqrt {{x^2} + 16} }}dx}  = \int {du}  = u + C = \ln \left| {x + \sqrt {{x^2} + 16} } \right| + C.$$
