How do you integrate ln(x+1)?

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Answer 1 · 2015-08-28T17:18:54

Notice how you can write this as:

#int 1*ln(x+1)dx#

To integrate this, you can do Integration by Parts.

Let:
#u = ln(x+1)#
#du = 1/(x+1)dx#
#dv = 1dx#
#v = x#

#uv - intvdu#

#= xln(x+1) - intx/(x+1)dx#

#= xln(x+1) - int(x+1-1)/(x+1)dx#

#= xln(x+1) - int1-1/(x+1)dx#

#= xln(x+1) - (x-ln(x+1))#

#= xln(x+1) + ln(x+1) - x#

#= color(blue)((x+1)ln(x+1) - x + C)#