Answer
$$
\begin{aligned}
g(x)&=\int_{2 x}^{3 x} \frac{u^{2}-1}{u^{2}+1} d u\\
&=\int_{2x}^{0} \frac{u^{2}-1}{u^{2}+1} d u+\int_{0}^{3 x} \frac{u^{2}-1}{u^{2}+1} d u\\
& =-\int_{0}^{2 x} \frac{u^{2}-1}{u^{2}+1} d u+\int_{0}^{3 x} \frac{u^{2}-1}{u^{2}+1} d u
\end{aligned}
$$
the derivative of the function $g(x)$ is given by the following:
$$
\begin{aligned}
g^{\prime}(x) &=-\frac{(2 x)^{2}-1}{(2 x)^{2}+1} \cdot \frac{d}{d x}(2 x)+\frac{(3 x)^{2}-1}{(3 x)^{2}+1} \cdot \frac{d}{d x}(3 x) \\
&=-2 \cdot \frac{4 x^{2}-1}{4 x^{2}+1}+3 \cdot \frac{9 x^{2}-1}{9 x^{2}+1}
\end{aligned}
$$
Work Step by Step
$$
\begin{aligned}
g(x)&=\int_{2 x}^{3 x} \frac{u^{2}-1}{u^{2}+1} d u\\
&=\int_{2x}^{0} \frac{u^{2}-1}{u^{2}+1} d u+\int_{0}^{3 x} \frac{u^{2}-1}{u^{2}+1} d u\\
& =-\int_{0}^{2 x} \frac{u^{2}-1}{u^{2}+1} d u+\int_{0}^{3 x} \frac{u^{2}-1}{u^{2}+1} d u
\end{aligned}
$$
the derivative of the function $g(x)$ is given by the following:
$$
\begin{aligned}
g^{\prime}(x) &=-\frac{(2 x)^{2}-1}{(2 x)^{2}+1} \cdot \frac{d}{d x}(2 x)+\frac{(3 x)^{2}-1}{(3 x)^{2}+1} \cdot \frac{d}{d x}(3 x) \\
&=-2 \cdot \frac{4 x^{2}-1}{4 x^{2}+1}+3 \cdot \frac{9 x^{2}-1}{9 x^{2}+1}
\end{aligned}
$$
