The interval in which tie function $f(x)=\int\limits_0^x\left(\frac{t}{t+2}-\frac{1}{t}\right) d t$ will be non-increasing, is

$(-2,-1) \cup(0,2)$

We have,

$f(x)=\int\limits_0^x\left(\frac{t}{t+2}-\frac{1}{t}\right) d t$

$\Rightarrow f'(x)=\frac{x}{x+2}-\frac{1}{x}=\frac{x^2-x-2}{x(x+2)}=\frac{(x-2)(x+1)}{x(x+2)}$

For f(x) to be non-increasing, we must have

$f'(x)<0 \Rightarrow \frac{(x-2)(x+1)}{x(x+2)}<0 \Rightarrow x \in(-2,-1) \cup(0,2)$