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

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Answer 1 · 2016-01-04T12:33:49

Basically you are asking for this

$int ln(x+1)dx=int x'*ln(x+1)dx=x*lnx(x+1)-int x*1/(x+1)dx= x*ln(x+1)-int(x+1-1)/(x+1)dx=x*ln(x+1)-x+ln(x+1)$

So the antiderivative of $ln(x+1)$ is

$x*ln(x+1)-x+ln(x+1)+c$

where c is a constant

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