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

1 Answer
bp
Sep 7, 2015

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

Explanation:

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

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