How do you find the integral of #5/[sqrt(9x^2-16)] dx#?

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Answer 1 · 2018-04-01T17:10:27

The answer is #=5/3ln(|3/4x+sqrt((3/4x)^2-1)|)+C#

Perform some simplification

#5/sqrt(9x^2-16)=5/(4sqrt (((3/4x)^2)-1))#

Let #3/4x=secu#, #=>#, #3/4dx=secutanudu#

#sqrt((3/4x)^2-1)=sqrt(sec^2u-1)=tanu#

Therefore, the integral is

#I=int(5dx)/(sqrt(9x^2-16))=5/4int(4/3secutanudu)/(tanu)#

#=5/3intsecudu#

#=5/3int(secu(secu+tanu)du)/(secu+tanu)#

Let #v=secu+tanu#, #=>#, #dv=(sec^2u+secutanu)du#

Therefore,

#I=5/3int(dv)/(v)#

#=5/3ln(v)#

#=5/3ln(secu+tanu)#

#=5/3ln(|3/4x+sqrt((3/4x)^2-1)|)+C#