Well, this is a bit hard !
Let's #u = sqrt(t^2-9)#
#du = t/sqrt(t^2-9)dt#
#dt = sqrt(t^2-9)/tdu#
Integral become :
#=>intsqrt(t^2-9)/(t^4sqrt(t^2-9))du = int1/t^4du#
#1/t^4 = 1/(u^2+9)^2#
So now we have #=> int 1/(u^2+9)^2du#
can't do partial fraction so let's #u = 3tan(v)#
#=>du = 3/cos^2(v)dv#
#=>1/(u^2+9)^2 = 1/(9tan^2(v)+9)^2 = cos^4(v)/81#
So now we have
#=>intcos^4(v)/81*3/cos^2(v)dv= 3/81intcos^2(v)dv#
#=>3/162int1+cos(2v)dv = 1/54int1+cos(2v)dv#
#=> 1/54[v+1/2sin(2v)]+C#
Substitute back for #v = arctan(1/3u)#
#=>1/54[arctan(1/3u)+1/2sin(2arctan(1/3u))]#
Substitute again for #u =sqrt(t^2-9)#
#=>1/54[arctan(1/3sqrt(t^2-9))+1/2sin(2arctan(1/3sqrt(t^2-9)))]#
