First, we make the substitution #u = e^x implies du = e^xdx#
Integral becomes:
#int (du)/(sqrt(u^2-81))#
Now, whatever trig substitution we use, we want the coefficient to be 9. This is because squaring 9 gives 81 which means we can then move a factor of 9 outside the square root.
Let #u = 9sec(theta) implies du = 9sec(theta)tan(theta)d theta#
#int (9sec(theta)tan(theta))/(sqrt(81sec^2(theta)-81))d theta#
#int (cancel(9)sec(theta)tan(theta))/(cancel(9)sqrt(sec^2(theta)-1))d theta = int (sec(theta)tan(theta))/(sqrt(sec^2(theta)-1))d theta#
We know the first Pythagorean identity:
#sin^2phi + cos^2phi = 1#
Dividing every term by #cos^2phi# gives the second identity:
#tan^2phi + 1 = sec^2phi implies tan^2phi = sec^2phi - 1#
So our integral becomes:
#int (sec(theta)tan(theta))/(sqrt(tan^2(theta)))d theta =int sec(theta)d theta #
This is a standard integral but does require some tricks to do if you don't have a table of integrals. It's only tangentially related to the question at hand though so I'll leave this link which has a very nice method of solving it.
#int sec(theta) d theta = ln(tan(theta) + sec(theta)) + C#
#u = 9sec(theta) implies theta = sec^(-1)(u/9)#
#=ln(tan(sec^(-1)(u/9)) + sec(sec^(-1)(u/9))) + C#
Now, to tackle #tan(sec^(-1)(u/9))#. I always find it very helpful to draw a triangle for this kind of thing but it's a bit time consuming so we'll use a quick and dirty method here..
Consider #y = sec^(-1)(x) implies x = sec(y)#
#therefore x^2 = sec^2(y)#
From the earlier pythagorean identity:
#x^2 = tan^2(y) + 1#
#x^2-1 = tan^2(y)#
#sqrt(x^2-1) = tan(y) = tan(sec^(-1)(x))#
So, our answer becomes:
#=ln(sqrt((u/9)^2-1) + u/9) + C#
#=ln(1/9(sqrt(u^2 - 81) + u)) + C#
#=ln(1/9(sqrt(e^(2x) - 81) + e^x)) + C#
