How do you integrate #int sqrt(-x^2-6x-18)/xdx# using trigonometric substitution?

1 Answer
Mar 28, 2017

Complete the square under the #sqrt#.

#int sqrt(-(x^2 + 6x + 9 - 9) - 18)/xdx#

#int sqrt(-(x^2 + 6x + 9) + 9 - 18)/xdx#

#int sqrt(-(x + 3)^2 - 9)/xdx#

Let #u = x + 3#. Then #du = dx#.

#int sqrt(-u^2 -9)/(u - 3) du#

But the expression under the #sqrt# clearly has no real values of #x# that satisfy. So techniqually this problem has no solution.

Hopefully this helps!