Substitute x =1/2 sect with t in (0,pi/2), dx = 1/2 sect tant dt:
int sqrt(20x^2-5)dx = 1/2 int sqrt(20( sect/2)^2-5)sect tant dt
int sqrt(20x^2-5)dx = 1/2 int sqrt(5 sec^2t-5)sect tant dt
int sqrt(20x^2-5)dx = sqrt5/2 int sqrt( sec^2t-1)sect tant dt
Use now the trigonometric identity:
sec^2t-1 = tant^2t
and as tanx > 0 for t in (0,pi/2):
sqrt(sec^2t-1) = tant
Then:
int sqrt(20x^2-5)dx = sqrt5/2 int sect tan^2t dt
and using the same identity:
int sqrt(20x^2-5)dx = sqrt5/2 int sect (sec^2t-1) dt
int sqrt(20x^2-5)dx = sqrt5/2 int sec^3tdt - sqrt5/2 int sectdt
Solve now the integrals:
int sect = ln abs (sect+tant)+C
and:
int sec^3tdt = int sectd/dt(tant)dt
Integrate by parts:
int sec^3tdt = sect tant - int tant d/dt(sect)dt
int sec^3tdt = sect tant - int sect tan^2t dt
int sec^3tdt = sect tant - int sect (sec^2t-1) dt
int sec^3tdt = sect tant - int sec^3tdt +int sectdt
Solve now for the original integral:
2int sec^3tdt = sect tant +int sectdt
int sec^3tdt = (sect tant)/2 +1/2 int sectdt
int sec^3tdt = (sect tant)/2 +1/2 ln abs (sect+tant) +C
Then:
int sqrt(20x^2-5)dx = sqrt5/4 sect tant - sqrt5/4 ln abs (sect+tant) +C
Undo the substitution:
tant = sqrt(sec^2t-1) = sqrt(4x^2-1)
sect = 2x
so:
int sqrt(20x^2-5)dx = sqrt5/2 xsqrt(4x^2-1) - sqrt5/4 ln (2absx+sqrt(4x^2-1)) +C