Completing the square at the denominator gives
#e^{2x} + 12e^x + 27 -= (e^x + 6)^2 - 9#
To make use of the identity
#sec^2u - 1 -= tan^2u#,
substitute #3secu = e^x + 6#.
Differentiate both sides w.r.t. #x# to get
#3secu tanu frac{du}{dx} = e^x#
#= 3secu - 6#
#frac{du}{dx} = frac{3secu - 6}{3secu tanu}#
#= frac{1 - 2cosu}{tanu}#
Now plugging it into the integration,
#int frac{1}{sqrt{(e^x + 6)^2 - 9}} dx = int frac{1}{sqrt{(3secu)^2 - 9}} frac{tanu}{1 - 2cosu} du#
#= 1/3 int frac{du}{1 - 2cosu}#
As #cosu -= frac{1-tan^2(u/2)}{1+tan^2(u/2)}#
Let #v = tan(u/2)#
Differentiating both sides w.r.t. #u#,
#frac{dv}{du} = 1/2 sec^2(u/2)#
#= ( 1 + tan^2(u/2) )/2#
#= frac{1 + v^2}{2}#
Plugging it into the integration,
#1/3 int frac{du}{1 - 2cosu} = 1/3 int frac{ frac{2dv}{1 + v^2} }{1-2(frac{1 - v^2}{1 + v^2})}#
#= 1/3 int frac{2dv}{(1 + v^2) - 2(1 - v^2)}#
#= 1/3 int frac{2dv}{3v^2 - 1}#
#= 1/3 int (frac{1}{sqrt{3}v - 1} - frac{1}{sqrt{3}v + 1}) dv#
#= frac{1}{3sqrt{3}} int (frac{sqrt{3}}{sqrt{3}v - 1} - frac{sqrt{3}}{sqrt{3}v + 1}) dv#
#= frac{1}{3sqrt{3}} ln(frac{sqrt{3}v - 1}{sqrt{3}v + 1}) + C_1#, where #C_1# is the constant of integration.
#= frac{1}{3sqrt{3}} ln(frac{tan(u/2) - 1/sqrt{3}}{tan(u/2) + 1/sqrt{3}}) + C_1#
As #tan^2(u/2) -= frac{secu - 1}{ secu + 1}#
#frac{1}{3sqrt{3}} ln(frac{tan(u/2) - 1/sqrt{3}}{tan(u/2) + 1/sqrt{3}}) + C_1 = frac{1}{3sqrt{3}} ln(frac{sqrt{frac{secu - 1}{secu + 1}} - 1/sqrt{3}}{sqrt{frac{secu - 1}{secu + 1}} + 1/sqrt{3}}) + C_1#
#= frac{1}{3sqrt{3}} ln(frac{sqrt{frac{frac{e^x + 6}{3} - 1}{frac{e^x + 6}{3} + 1}} - 1/sqrt{3}}{sqrt{frac{frac{e^x + 6}{3} - 1}{frac{e^x + 6}{3} + 1}} + 1/sqrt{3}}) + C_1#
#= frac{1}{3sqrt{3}} ln(frac{sqrt{frac{e^x + 3}{e^x + 9}} - 1/sqrt{3}}{sqrt{frac{e^x + 3}{e^x + 9}} + 1/sqrt{3}}) + C_1#
#= frac{1}{3sqrt{3}} ln(frac{sqrt{3}sqrt{e^x + 3} - sqrt{e^x + 9}}{sqrt{3}sqrt{e^x + 3} + sqrt{e^x + 9}}) + C_1#
#= frac{1}{3sqrt{3}} ln(frac{(sqrt{3}sqrt{e^x + 3} - sqrt{e^x + 9})^2}{3(e^x + 3) - (e^x + 9)}) + C_1#
#= frac{1}{3sqrt{3}} ln(frac{(sqrt{3}sqrt{e^x + 3} - sqrt{e^x + 9})^2}{2e^x }) + C_1#
#= frac{1}{3sqrt{3}} ln((sqrt{3}sqrt{e^x + 3} - sqrt{e^x + 9})^2) - frac{1}{3sqrt{3}} ln(2e^x) + C_1#
#= frac{2}{3sqrt{3}} ln(sqrt{3}sqrt{e^x + 3} - sqrt{e^x + 9}) - frac{x}{3sqrt{3}} + C_2#, where #C_2 = C_1 - frac{ln2}{3sqrt{3}}#
#= frac{2ln(sqrt{3e^x + 9} - sqrt{e^x + 9})-x}{3sqrt{3}} + C_2#
