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

$uv - intvdu$

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

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

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

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

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

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