integration by parts

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s- integral

s ln (2x+1)dx ?

= ln(2x+1)x - s x d( ln (2x+1))
= ln(2x+1)x- s x [(2x+1)'/ (2x+1)] dx
= ln(2x+1)x- s x [(2)/ (2x+1)] dx
= ln(2x+1)x- s (2x/2x+1) dx
= ln(2x+1)x- s [(2x+1-1)/(2x+1)] ... i'm confused about why there is a minus 1 after the integral

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