#e^x=4sinΘ#
#x=ln(4sinΘ)#
#dx=1/(4sinΘ)*4cosΘdΘ#
#dx=cosΘ/sinΘdΘ#
#int(e^x-1)/(sqrt(e^(2x)-16)##dx#
#int(4sinΘ-1)/(sqrt((4sinΘ)^2-16)##*cosΘ/sinΘdΘ#
#int(4sinΘ-1)/(sqrt(16sin^2Θ-16))##*cosΘ/sinΘdΘ#
#int(4sinΘ-1)/(sqrt(16)*sqrt(sin^2Θ-1)##*cosΘ/sinΘdΘ#
#int(4sinΘ-1)/(4*sqrt(-cos^2Θ)##*cosΘ/sinΘdΘ#
#1/iint(4sinΘ-1)/(4cosΘ##*cosΘ/sinΘdΘ#
#-iint(4sinΘ-1)/(4sinΘ)dΘ#
#-iint((4sinΘ)/(4sinΘ)-(1)/(4sinΘ))dΘ#
#-iint1dΘ+iint1/4cscΘdΘ#
#-iΘ + i/4ln|cscΘ+cotΘ| + c#
#color(red)( bar( ul( | color(white)(a/a) color(black)( -iarcsin(e^x/4)-i/4ln(|(4+sqrt(16-e^(2x)))/e^x|)+c, c in RR ) color(white)(a/a) | )))#
Note 1: We set #e^x# as a form of #sinΘ# because the #e^x# term in the root in the denominator was positive and the constant term was negative.
Note 2: #intcscΘ#=#-ln|cscx+cotx|+c#
Note 3: You get the expression for #Θ# from rewriting the original equation, and you can solve for #cscΘ# and #cotΘ# using a right triangle.
Note 4: There are probably ways to simplify my final answer, but I just kept it like that.