19: The domain is D( f ) = (−1,0) ∪ (0,∞).
The derivative is

Critical points: The equation f ′(x) = 0 cannot be solved analytically. Since the denomintor is always positive in the domain, the sign of f ′ is determined by its numerator

g(x) = x − (x + 1)⋅ln(x + 1).

We can guess that g = 0 for x = 0. However, we still cannot handle g analytically. Trick: We check on how g goes. We find g′(x) = −ln(x + 1), its signs tell us that g is increasing on (−1,0) and decreasing on (0,∞). This shows that g(0) = 0 is a local maximum and also a global maximum and so g is never positive for x > −1.

We just learned that there are no critical points, so the intervals of monotonicity will be (−1,0) and (0,∞), and also that f ′ < 0 there. This should make the monotonicity chart obvious.

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