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Answer 1 · 2018-01-16T10:54:59

#int\ dx/sqrt(-x^2-8x)=arcsin((x+4)/4)+C#

#\ \ \ \ \ \ int\ dx/sqrt(-x^2-8x)#

Completing the square, we have
#=int\ dx/sqrt(16-(x+4)^2)#

Substitute #u=x+4# and #du=dx#:
#=int\ (du)/sqrt(16-u^2)#

Substitute #u=4sin(theta)# and #du=4cos(theta)\ d theta#:
#=int\ (4cos(theta))/sqrt(16-16sin^2(theta))\ d theta#

#=int\ (4cos(theta))/(4cos(theta))\ d theta#

#=int\ d theta#

#=theta+C#

#=arcsin(u/4)+C#

#=arcsin((x+4)/4)+C#