# 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+1-1)/(2x+1)] ... i'm confused about why there is a minus 1 after the integral