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:

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

To integrate this, you can do Integration by Parts.

Let:
$u = ln (x + 1)$ $u = ln(x+1)$
$d u = \frac{1}{x + 1} d x$ $du = 1/(x+1)dx$
$d v = 1 d x$ $dv = 1dx$
$v = x$ $v = x$

$u v - \int v d u$ $uv - intvdu$

$= x ln (x + 1) - \int \frac{x}{x + 1} d x$ $= xln(x+1) - intx/(x+1)dx$

$= x ln (x + 1) - \int \frac{x + 1 - 1}{x + 1} d x$ $= xln(x+1) - int(x+1-1)/(x+1)dx$

$= x ln (x + 1) - \int 1 - \frac{1}{x + 1} d x$ $= xln(x+1) - int1-1/(x+1)dx$

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

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

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