For integrals involving the root #sqrt(x^2+a^2),# we can use the substitution #x=atantheta#. Here, #a^2=16, a=4, x=4tantheta#
So, #dx=4sec^2thetad theta# and we can apply the substitution.
#int(4sec^2theta)/sqrt(16+16tan^2theta)d theta=int(4sec^2theta)/(sqrt(16(1+tan^2theta)))d theta#
#=intsec^2theta/sqrt(1+tan^2theta)d theta#
Recalling the identity #1+tan^2theta=sec^2theta,#
#=intsec^2theta/sqrt(sec^2theta)d theta#
#=intsec^cancel(2)theta/cancel(secthetad) theta#
#=intsecthetad theta=ln|sectheta+tantheta|+C#
This is a fairly common integral, it should be memorized. We need things in terms of #x.# Recalling that #x=4tantheta, tantheta=x/4.#
We still need the secant in terms of #x.# Applying the identity #1+tan^2theta=sec^2theta:#
#16/16+x^2/16=sec^2theta#
#sectheta=sqrt(x^2+16)/4#
Thus, we have
#ln|(sqrt(x^2+16)+x)/4|+C=ln|sqrt(x^2+16)+x|-ln(4)+C#
We may absorb the #ln4# into the constant of integration. We're left with
#intx/sqrt(16+x^2)dx=ln|sqrt(x^2+16)+x|+C#
