Let #x=6sectheta#. This means that:
#dx=6secthetatantheta# #d theta#
Performing the substitution:
#color(white)=int 1/(x^2sqrt(x^2-36))dx#
#=int1/((6sectheta)^2sqrt((6sectheta)^2-36))6secthetatantheta# #d theta#
#=int1/((6sectheta)^color(red)cancelcolor(black)2sqrt((6sectheta)^2-36))color(red)cancelcolor(black)(6sectheta)tantheta# #d theta#
#=int1/(6secthetasqrt((6sectheta)^2-36))tantheta# #d theta#
#=int1/(6secthetasqrt(36sec^2theta-36))tantheta# #d theta#
#=int1/(6secthetasqrt(36(sec^2theta-1)))tantheta# #d theta#
#=int1/(6sectheta*6sqrt(sec^2theta-1))tantheta# #d theta#
By the Pythagorean identity #tan^2theta+1=sec^2theta#:
#=int1/(36sectheta*sqrt(tan^2theta+1-1))tantheta# #d theta#
#=int1/(36sectheta*sqrt(tan^2theta))tantheta# #d theta#
#=int1/(36sectheta*tantheta)tantheta# #d theta#
#=int1/(36sectheta*color(red)cancelcolor(black)tantheta)color(red)cancelcolor(black)tantheta# #d theta#
#=int1/(36sectheta)# #d theta#
#=1/36int1/sectheta# #d theta#
#=1/36int1/(1/costheta)# #d theta#
#=1/36intcostheta# #d theta#
#=1/36sintheta+C#
Now we want #sintheta# in terms of #x#. By our substitution earlier, we know:
#6sectheta=x#
#sectheta=x/6#
#1/costheta=x/6#
#costheta=6/x#
#cos^2theta=36/x^2#
#-cos^2theta=-36/x^2#
#-cos^2theta+1=-36/x^2+1#
#1-cos^2theta=1-36/x^2#
#1-cos^2theta=(x^2-36)/x^2#
#sin^2theta=(x^2-36)/x^2#
#sintheta=sqrt((x^2-36)/x^2)#
#sintheta=sqrt(x^2-36)/x#
Plugging this back into the integral:
#color(white)=1/36sintheta+C#
#=1/36*sqrt(x^2-36)/x+C#
#=sqrt(x^2-36)/(36x)+C#
That's the whole integral. Hope this helped!
