#int (1)/(sqrt(4x^2-9))dx=#?

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Answer 1 · 2018-02-17T01:40:54

#1/2ln abs(x+sqrt(x^2-9/4))+C#

We'll use the following integration form:
#int (1) /(sqrt(u^2 - a^2))du=ln abs(u + sqrt(u^2 - a^2))+C#

So:

#int (1)/(sqrt(4x^2-9))dx#

#=int (1)/(sqrt(4(x^2-9/4)))dx#

#=1/2 int 1/(sqrt(x^2-9/4)) dx#

#=1/2ln abs(x+sqrt(x^2-9/4))+C#

The function's domain is #{x : sqrt(x^2-9/4)>0}#, which reduces to #x in (–oo, –3/2) uu (3/2, oo).#