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,

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

$=5/3ln(v)$

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

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

How do you find the integral of 5/[sqrt(9x^2-16)] dx5/[sqrt(9x^2-16)] dx?