The solution is really quite long but there is a solution;
start with a Right triangle.
with acute angle #theta#
side opposite to angle #theta# = (x-4)
side opposite to 90 degree angle= #sqrt(x^2-8x+32)#
side adjacent to #theta# = 4
Our trigonometric substitution goes like this:
Let #x-4 = 4*tan theta#
and #x=4+4*tan theta#
also #(x-4)^2 = 16 *tan^2 theta#
#dx = 4* sec^2 theta# #d##theta#
and #tan theta = (x-4)/4#
our #sec theta = sqrt(x^2 -8x+32)/4#
Take note that #sqrt(x^2-8x+32) = sqrt((x-4)^2+16)#
from the given:
#intx/sqrt(-x^2+8x-32) dx#
#int x/sqrt(-(x^2-8x+16+16)) dx#
#int x/sqrt(-((x-4)^2+16)) dx#
from the above it follows
#int ((4+4*tan theta) * 4 sec^2 theta )/sqrt(-1(16*tan^2 theta+16)# #d##theta#
#int (4(1+tan theta) * 4 sec^2 theta )/(sqrt(-1) *sqrt(16*tan^2 theta+16))# #d##theta#
#int (4(1+tan theta) * 4 sec^2 theta )/(sqrt(-1) *sqrt(16)*sqrt(tan^2 theta+1))# #d##theta#
#int (4(1+tan theta) * sec^2 theta )/(sqrt(-1) *sqrt(tan^2 theta+1))# #d##theta#
but #sqrt(tan^2 theta+1) = sec theta#
#int (4(1+tan theta) * sec^2 theta )/(sqrt(-1) *sec theta)# #d##theta#
then #int (4(1+tan theta) * sec theta )/(sqrt(-1) )# #d##theta#
it follows #int (4(sec theta+tan theta sec theta) )/(sqrt(-1) )# #d##theta#
and #(4/sqrt(-1))int sec theta# #d##theta# + # (4/sqrt(-1)) int tan theta sec theta# #d##theta #
but #sqrt(-1) = i# and #i^2 = -1 #
#(4/sqrt(-1)*sqrt(-1)/sqrt(-1))int sec theta# #d##theta# + # (4/sqrt(-1)*sqrt(-1)/sqrt(-1)) int tan theta sec theta# #d##theta #
#((4i)/(-1))int sec theta# #d##theta# + # ((4i)/(-1)) int tan theta sec theta# #d##theta #
after integration using the basic formulas
#((4i)/-1)ln (sec theta + tan theta)# + # ((4i)/-1) sec theta #
returning to our original variables and simplifying, it follows,
#-4i* ln ((x-4+sqrt(x^2-8x+32))/4)-i*sqrt(x^2-8x+32)#
