Without trigonometric substitution :
Here ,
I=intx/sqrt(4x^2+4x+-24)dx
=1/2intx/sqrt(x^2+x+-6)dx
=1/4int(2x)/sqrt(x^2+x+-6)dx
=1/4int(2x+1-1)/sqrt(x^2+x+-6)dx
=1/4int(2x+1)/sqrt(x^2+x+-6)dx-1/4int1/sqrt(x^2+x+-6)dx
=1/4int(d(x^2+x+-6))/sqrt(x^2+x+-6)-1/4I_1
=1/4*2sqrt(x^2+x+-6)-1/4I_1
=1/2sqrt(x^2+x+-6)-1/4I_1.....................to(A)
Now ,
I_1=int1/sqrt(x^2+x+-6)dx
Now ,
(i)x^2+x+6=x^2+x+1/4+23/4=(x+1/2)^2+23/4
(ii)x^2+x-6=x^2+x+1/4-25/4=(x+1/2)^2-25/4
So ,
I_1=int1/sqrt((x+1/2)^2+K)dx ,where ,K=23/4 orK=-25/4
:.I_1=ln|(x+1/2)+sqrt((x+1/2)^2+K)|+c'
:.I_1=ln|(x+1/2)+sqrt(x^2+x+1/4+K)|+c'
substitute K=23/4 orK=-25/4
:.I_1=ln|x+1/2+sqrt(x^2+x+-6)|+c'
From (A) we get
I=1/2sqrt(x^2+x+-6)-1/4ln|x+1/2+sqrt(x^2+x+-6)|+C
Note :
color(red)(int(f'(x))/sqrt(f(x))dx=2sqrt(f(x))+c
OR
color(red)(int(d[f(x)])/sqrt(f(x))=2sqrt(f(x))+c