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!