The substitution $x\=\frac{2}{3}\sec \left(\mathrm{\θ}\right)$ means $\mathrm{dx}\=\frac{2}{3}\sec \left(\mathrm{\θ}\right)\tan \left(\mathrm{\θ}\right) d\mathrm{\θ}$, and turns $h\left(x\right)$ into $2 \tan \left(\mathrm{\θ}\right)$. From Figure 6.3.3, $\tan \left(\mathrm{\θ}\right)\=\frac{1}{2}\sqrt{9 x^2-4}$. Hence, the evaluation of the given integral proceeds as follows.

Passage from line 2 to line 3 is via the trig identity ${\tan }^2\left(\mathrm{\θ}\right)\={\sec }^2\left(\mathrm{\θ}\right)-1$. The antiderivative of ${\sec }^2\left(\mathrm{\θ}\right)$ is $\tan \left(\mathrm{\θ}\right)$ because the derivative of $\tan \left(\mathrm{\θ}\right)$ is ${\sec }^2\left(\mathrm{\θ}\right)$.

Finally, note that here, $\mathrm{\θ}\=\mathrm{arcsec}\left(\frac{3}{2}x\right)$ or $\mathrm{\θ}\=\arctan \left(\frac{\sqrt{9 x^2-4}}{2}\right)$.

A stepwise solution that uses top-level commands except for one application of the Change command from the IntegrationTools package:

From Figure 6.3.3, $\tan \left(\mathrm{\θ}\right)\=\frac{1}{2}\sqrt{9 x^2-4}$.

The stepwise solution provided by the  tutor when the Constant, Constant Multiple, and Sum rules are taken as Understood Rules begins with the substitution $u^2\=9 x^2-4$ and proceeds as shown in Table 6.3.21(a). The change of variables selected by the tutor leads to the rational function $u^2\/\left(u^2\+4\right)$, which immediately becomes $1-4\/\left(u^2\+4\right)$ by long division. The antiderivative of $1\/\left(u^2\+4\right)$ can be found by coercing this fraction into the form $1\/\left(z^2\+1\right)$, so that $\arctan \left(z\right)$ results. The tutor, however, makes the change of variables $u\=2 \tan \left(v\right)$, which converts the integrand to $1\/2$.

Table 6.3.21(b) shows the result when the Change rule $x\=\frac{2}{3}\sec \left(\mathrm{\θ}\right)$ is imposed on the tutor.

After making the initial change of variables, the integrand is actually $2 {\tan }^2\left(\mathrm{\θ}\right)$, which becomes $2 u^2\/\left(u^2\+1\right)$. As in Table 6.3.21(a), Maple immediately writes this rational function as $2\left(1-1\/\left(u^2\+1\right)\right)$. The further change of variables $u\=\tan \left(v\right)$ changes $\int \frac{\mathrm{du}}{u^2\+1}$ to $\int \mathrm{dv}$. Here, it is well to note that the integration rules are names for the properties or expressions in the integrand. Therefore, the arctan rule does not apply because the arctangent arises not in the integrand, but only after the integral is evaluated. That's why the additional change of variables is needed in both Tables 6.3.21(a) and 6.3.21(b).