Complete the square at the denominator:
#int dx/sqrt(4x^2+12x-5) = int dx/sqrt((2x+3)^2-14) = 1/sqrt14 int dx/sqrt(((2x+3)/sqrt14)^2-1)#
Substitute for #(2x+3)/sqrt 14 > 1#
#(2x+3)/sqrt14 = sect#
#dx = sqrt14/2 sect tantdt#
with #t in (0,pi/2)#
so that:
#int dx/sqrt(4x^2+12x-5) = 1/2 int (sect tant dt)/sqrt(sec^2t -1)#
Now:
#sqrt(sec^2t -1) = sqrt(1/cos^2t -1) = sqrt((1-cos^2t)/cos^2t) = sqrt(sin^2t/cos^2t) = sqrt(tan^2t) = tant#
as for #t in (0,pi/2)# the tangent is positive.
So:
#int dx/sqrt(4x^2+12x-5) = 1/2 int (sect tant dt)/tant #
#int dx/sqrt(4x^2+12x-5) = 1/2 int sect dt #
#int dx/sqrt(4x^2+12x-5) = 1/2 ln abs (sect+tant)+C#
#int dx/sqrt(4x^2+12x-5) = 1/2 ln abs (sect+sqrt(sec^2t-1))+C#
Undoing the substitution:
#int dx/sqrt(4x^2+12x-5) = 1/2 ln abs ((2x+3)/sqrt14+sqrt(((2x+3)/sqrt14)^2-1))+C#
#int dx/sqrt(4x^2+12x-5) = 1/2 ln abs (2x+3+sqrt(4x^2+12x-5))+C#
and by differentiating we can verify that the solution is valid also for #(2x+3)/sqrt14 < -1#.
