已知函数f（x）=ln(x+a)-x∧2-x在x＝0处取得极值,

f(x)=ln(x+a)-x^2-x
f'(x)=1/(x+a)-2x-1
因为x＝0处取得极值
则f'(0)=1/a-1=0
a=1
f'(x)=1/(x+1)-2x-1
=[1-2x(x+1)-(x+1)]/(x+1)
=-x(2x+3)/(x+1)=0
增区间是x=0