How do you integrate #int 1/sqrt(x^2-9x-7) # using trigonometric substitution?

For this, you're going to have to complete the square.

#x^2 - 9x - 7#

#7 = x^2 - 9x#

#28/4 + 81/4 = x^2 - 9x + 81/4#

#109/4 = (x - 9/2)^2#

Now bring it back into the problem.

#int 1/(sqrt((x - 9/2)^2 - 109/4))dx#

Let:
#u = x - 9/2#
#du = dx#

#= int 1/(sqrt(u^2 - 109/4))du#

Now it looks more like a #sqrt(x^2 - a^2)# form. If we let #a = sqrt(109)/2#, then:

#u = sqrt(109)/2sectheta#
#du = sqrt(109)/2secthetatanthetad theta#
#sqrt(u^2 - 109/4) = sqrt(109/4sec^2theta - 109/4) = sqrt(109)/2tantheta#

What we now have is:

#= int 1/(cancel(sqrt(109)/2)cancel(tantheta))*cancel(sqrt(109)/2)secthetacancel(tantheta)d theta#

#= int secthetad theta#

Remember the trick to do this? See the following:

#color(green)(intsecxdx)#

#= int (secx(secx+tanx))/(secx+tanx)dx#

#= int (sec^2x+secxtanx)/(secx+tanx)dx#

Let:
#u = secx+tanx#
#du = secxtanx+sec^2xdx#

#=> int 1/udu = ln|u| + C = color(blue)(ln|secx + tanx| + C)#

Therefore, what we have is #ln|sectheta + tantheta|# for the result. Now what we can do is substitute in the proper variables.

#tantheta = (2sqrt(u^2 - 109/4))/sqrt(109) = (2sqrt(x^2 - 9x - 7))/sqrt(109)#
#sectheta = (x - 9/2)*2/sqrt(109) = (2x - 9)/sqrt(109)#

So the answer is:

#= color(green)(ln|(2x - 9)/sqrt(109) + (2sqrt(x^2 - 9x - 7))/sqrt(109)| + C)#

or, embedding the #sqrt(109)# into the #C#:

#= ln|2x - 9 + 2sqrt(x^2 - 9x - 7)| - ln sqrt(109) + C#

#= color(blue)(ln|2x - 9 + 2sqrt(x^2 - 9x - 7)| + C)#