What is the antiderivative of $\frac{x}{x + 1}$ $x/(x+1)$ ?

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Answer 1 · 2015-07-07T02:22:25

$x + ln | x + 1 | + C$ $x + ln abs(x+1) +C$
Note: This is the same answer as the one Selim has given. It just has a different constant.

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

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

$= x + ln | x + 1 | + C$ $= x +ln abs(x+1) +C$

What is the antiderivative of x/(x+1)xx+1 x/(x+1)?