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

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Answer 1 · 2014-12-23T12:26:35

$\int u d v = d v - \int v d u$ $int udv=dv-int vdu$

$\int ln (x + 1) d x$ $int ln(x+1) dx$

by Substitution $t = x + 1 \Rightarrow d t = d x$ $t=x+1 => dt=dx$ ,

$= \int ln t d t$ $= int ln t dt$

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.

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