Calculus: Early Transcendentals, Chapter 7, 7.3, Section 7.3, Problem 9

intdx/sqrt(x^2+16)
Let x=4tantheta for -pi/2dx/[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.