`int (dx)/(sqrt(x^2 + 16))` Evaluate the integral

`intdx/sqrt(x^2+16)`

Let  `x=4tantheta`  for `-pi/2<theta<pi/2` . 

`dx/[d(theta)]=4sec^2theta`

` ` `dx=4sec^2thetad(theta)`

`sqrt(x^2+16)=`

`sqrt((4tantheta)^2+16)=`

`sqrt(16tan^2theta+16)=`

`sqrt[16(tan^2theta+1)]=`

`sqrt(16sec^2theta)=`

`4|sec(theta)|`

`intdx/[4sec(theta)]=`

`int(4sec^2(theta)d(theta))/(4sec(theta))=`

` ` `intsec(theta)d(theta)=`

`ln|sec(theta)+tan(theta)|+C_1=`

`ln|sqrt(x^2+16)/4+x/4|+C_1=`

`ln|(sqrt(x^2+16)+x)/4|+C_1=`

`ln|(sqrt(x^2+16)+x)|-ln4+C_1=`

`ln|sqrt(x^2+16)+x|+C`

where `C`  is the constant `C_1-ln4` .

The final answer is 

`ln|(sqrt(x^2+16)+x)+C`

where `C` is the constant `C_1-ln4`.