How do I evaluate #intsqrt(4x^2-9)/xdx#?

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Answer 1 · 2018-06-22T08:26:44

The answer is #=sqrt(4x^2-9)-3arctan(1/3sqrt(4x^2-9))+C#

Let #u=sqrt(4x^2-9)#, then

#du=(8xdx)/(2sqrt(4x^2-9))=(4xdx)/sqrt(4x^2-9)#

The integral is

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

#=int(dusqrt(4x^2-9)*sqrt(4x^2-x))/(4x*x)#

#=int(u^2du)/(u^2+9)#

#=int((u^2+9)du)/(u^2+9)-int(9du)/(u^2+9)#

#=intdu-9int(du)/(u^2+9)#

#=u-9int(du)/(u^2+9)#

#I_1-I_2#

The second integral is

#I_2=9int(du)/(u^2+9)#

Let #v=u/3#, #=>#, #dv=(du)/3#

#I_2=9int(3dv)/(9v^2+9)#

#=27/9int(dv)/(v^2+1)#

#=3arctan(v)#

#=3arctan(u/3)#

Finally the integral is

#I=u-3arctan(u/3)#

#=sqrt(4x^2-9)-3arctan(1/3sqrt(4x^2-9))+C#