Y=lnx-2/(x- 1)- 1, because y = lnx and y=-2/(x- 1) are monotonically increasing functions, this function is monotonically increasing.

However, because y=-2/(x- 1) is meaningless when x= 1, the function monotonically increases in the region of (0, 1, ∞).

PS: note, you can't write union! Because when this function is at x= 1, the function value has changed greatly! The left infinity of 1 is close to negative infinity, the right infinity of 1 is close to positive infinity, the function value is discontinuous and negative, and the infinite moment becomes positive infinity), so you can't write union, you can only write "and" or "and".

4. Classification discussion: When a=- 1, y =-x 2+1,it increases monotonically at (-∞, 0) and decreases monotonically at (0,+∞).

A< is at-1, and the coefficient before lnx is less than 0, so (a+ 1)lnx monotonically decreases, because the definition domain at this time is (0, +∞) and ax 2+1monotonically decreases at (0, +∞), so the function monotonically decreases.

A>0, the function is monotonically increasing, and the definition domain and analysis method are the same as A.

When a ∈ (- 1, 0), (a+ 1)lnx monotonically increases and ax 2+1monotonically decreases, we take the derivative.

Y'=(a+ 1)/x+2ax, so y' = 0 (both sides are multiplied by x simultaneously).

X = root number (-0.5-0.5/a), because a ∈ (- 1 0) can be regarded as (-0.5-0.5/a)>0, so this X exists (the number under the root number must be greater than or equal to 0).

Define the domain (0, +∞) to make the function monotonically increase at (0, x) and monotonically decrease at (x, +∞).