What is the integral of #ln(x+1)#?

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Answer 1 · 2014-12-23T12:26:35

#int udv=dv-int vdu#

#int ln(x+1) dx#

by Substitution #t=x+1 => dt=dx#,

#= int ln t dt#

by Integration by Parts with #{(u=ln t => du ={dt}/t),(dv=dt => v=t):}#,

#=t ln t - int dt#

by evaluating the integral,

#=t ln t - t+C#

by putting #t=x+1# back in,

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

(Note: You may also let the constant #C# include the last #+1#.)

I hope that this was helpful.