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September 16, 2014

September 16, 2014

Posted by **nicholas** on Saturday, February 17, 2007 at 9:01pm.

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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