Let, #I=intx^2/sqrt(x^2-4)dx#.
We subst. #x=2secy," so that, "dx=2secytanydy#.
#:. I=int(4sec^2y)/sqrt(4sec^2y-4)*2secytanydy#,
#=int(4sec^2y)/(2tany)*2secytanydy#,
#=4intsec^3ydy#,
#=4intsecy*sec^2ydy#,
#=4intsqrt(tan^2y+1)sec^2ydy#.
Next, we use the substn. #tany=t :. sec^2ydy=dt#.
#:. I=4intsqrt(t^2+1)dt#,
#=4{t/2sqrt(t^2+1)+1/2ln|t+sqrt(t^2+1)|}#,
#=2{tanysecy+ln|tany+secy|}......[because, t=tany]#.
Returning to #secy=x/2," so that, "tany=sqrt(x^2/4-1)#, we have,
#I=2{x/2*sqrt(x^2-4)/2+ln|x/2+sqrt(x^2-4)/2|}#,
#rArr I=x/2sqrt(x^2-4)+2ln|x/2+sqrt(x^2-4)/2|+C, or, #
# I=x/2sqrt(x^2-4)+2ln|x+sqrt(x^2-4)|+c, c=C-2ln2#.
However, the integral can easily be dealt with without
using the substn. as shown below :
#I=intx^2/sqrt(x^2-4)dx=int{(x^2-4)+4}/sqrt(x^2-4)dx#,
#int{(x^2-4)/sqrt(x^2-4)+4/sqrt(x^2-4)}dx#,
#=intsqrt(x^2-4)dx+4int1/sqrt(x^2-4)dx#,
#={x/2sqrt(x^2-4)-4/2ln|x+sqrt(x^2-4)|}+4ln|x+sqrt(x^2-4)|#.
#I=x/2sqrt(x^2-4)+2ln|x+sqrt(x^2-4)|+C_1#, as before!
Enjoy Maths.!