integration by parts
posted by nicholas .
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+11)/(2x+1)] ... i'm confused about why there is a minus 1 after the integral
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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)] ? 
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