How do you find the anti derivative of $(sqrtx -8)/(sqrtx-4)$ ?

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Answer 1 · 2015-09-07T07:05:32

$int (sqrtx -8)/(sqrtx -4) dx=x-8sqrtx -32ln|sqrtx -4|$ +C

The expression is equivalent to $1 -4/(sqrtx -4)$

$int (sqrtx -8)/(sqrtx -4) dx$ = $int (1 -4/(sqrtx -4) )dx= x -4int 1/(sqrtx -4) dx$

Now to integrate $1/(sqrtx -4) dx$ , let x =t^2, so that dx= 2tdt and so $int 1/(sqrtx-4) dx= int (2tdt)/(t-4)$

= $2int 1+4/(t-4) dt$ = 2(t+4 ln|t-4))= $2sqrtx +8ln|sqrtx-4|$

$int (sqrtx -8)/(sqrtx -4) dx=x-8sqrtx -32ln|sqrtx -4|$ +C

How do you find the anti derivative of (sqrtx -8)/(sqrtx-4)(sqrtx -8)/(sqrtx-4)?