Derivative of the definite integral I have to find the derivative for
$$\int_{-1}^x \frac{t^2}{t^2+4}dt-\int_{3}^x \frac{t^2}{t^2+4}dt$$
When I calculate their derivative separately
$$\frac {d}{dt} \left(\int_{-1}^x \frac{t^2}{t^2+4}dt\right)-\frac {d}{dt}\left(\int_{3}^x \frac{t^2}{t^2+4}dt\right)$$
it gives me the right result, that is $0$.
But according to the law of additive
\begin{aligned}
&\int_{-1}^x \frac{t^2}{t^2+4}dt-\int_{3}^x \frac{t^2}{t^2+4}dt \\
= & \int_{-1}^x \frac{t^2}{t^2+4}dt+\int_{x}^3 \frac{t^2}{t^2+4}dt \\= &\int_{-1}^3 \frac{t^2}{t^2+4}dt \\[2em]
\end{aligned}
$$d/dt\int_{-1}^3 \frac{t^2}{t^2+4}dt = \frac{3^2}{3^2+4} = \frac 9{13}$$
Thanks.

You have to obtain the derivative $\frac d{dx}$ of $$\int_{-1}^3 \frac{t^2}{t^2+4}dt$$
so the result will be
$$\frac d{dx}\int_{-1}^3 \frac{t^2}{t^2+4}dt =\frac d{dx}(const.) = 0$$
