#int (1)/(sqrt(4x^2-9))dx=#?
1 Answer
Feb 17, 2018
Explanation:
We'll use the following integration form:
So:
#int (1)/(sqrt(4x^2-9))dx#
The function's domain is
We'll use the following integration form:
So:
#int (1)/(sqrt(4x^2-9))dx#
The function's domain is