LOL. Yeah I thought there was an x there.
integral (square root of (16 + x^2)
t=x+sqrt(x^2+16)
(t-x)^2=x^2+16
t^2-2xt+x^2=x^2+16
t^2-16=2xt
x=(t^2-16)/2t
So sqrt(x^2+16)=t-x=(t^2+16)/2t
dx=(t^2+16)/2t^2 dt
∫ sqrt(x^2+16) dx=1/4 ∫ (t^2+16)^2/t^3 dt
1/4 ∫ (t^4+32t^2+256)/t^3 dt
1/4 ∫ t+32/t+256t^(-3) dt
1/4(t^2/2+32ln(t)-256/2t^2) + C
Now,
t=x+sqrt(x^2+16)
t^2=x^2+2sqrt(x^2+16)+x^2+16
t^2=2x^2+16+2sqrt(x^2+16)
256/t^2=2x^2+16-2xsqrt(x^2+16) (rationalising)
So,
∫ sqrt(x^2+16)dx = 1/4(t^2/2+32ln(t)-256/2t^2) =
1/8(2x^2+16-2xsqrt(x^2+16)+2x^2+16+2xsqrt(x^2+16)+8ln(x+sqrt(x^2+16) + C =
(xsqrt(x^2+16)+16ln(x+sqrt(x^2+16))/2 + C