Instead of adding up the estimates for the local errors, we can use the standard expression for the global error, which in general is $-\frac{(b-a)^3}{12N^2}f''(c),$ where $c$ is a number between $a$ and $b$. In our case, $a=0$ and $b=1$. The number $N$ of subintervals is $200$.
Now we need an estimate for $f''(c)$. The second derivative is, I think, $\dfrac{2(1-x^2)}{(1+x^2)^2}$. As $x$ increases, the numerator is decreasing, and the denominator is increasing. Thus $f''(x)$ is steadily decreasing. It is equal to $2$ at $x=0$, so $0\le f''(x)\lt 2$.
Remark: Or else use the local errors, and add them up. Since the second derivative is everywhere positive and $\lt 2$, the local error is always negative and $\lt \dfrac{1}{12}\cdot\dfrac{1}{(200)^3}\cdot 2$ in absolute value, so in absolute value the total error is $\le 200\cdot \dfrac{1}{12}\cdot\dfrac{1}{(200)^3}\cdot 2$.
Typically, the error estimates produced by this kind of calculation are unduly pessimistic. One can ordinarily get more useful information by comparing $T_{100}$ with $T_{200}$. The number $T_{100}$ is typically computed on the way to computing $T_{200}$, so it involves no additional cost.