The substitution $x\=\frac{2}{3}\tan \left(\mathrm{\θ}\right)$ means $\mathrm{dx}\=\frac{2}{3}{\sec }^2\left(\mathrm{\θ}\right) d\mathrm{\θ}$, and turns $g\left(x\right)$ into $2 \sec \left(\mathrm{\θ}\right)$. From Figure 6.3.2, $\sec \left(\mathrm{\θ}\right)\=\frac{1}{2}\sqrt{4\+9 x^2}$ and $\cos \left(\mathrm{\θ}\right)\=2\/\sqrt{4\+9 x^2}$. Hence, the evaluation of the given integral proceeds as follows.

The identity $\frac{1}{u^2\left(1-u^2\right)}\=\frac{1}{1-u^2}\+\frac{1}{u^2}$ is established by the algebraic device of partial fraction decomposition that will be studied formally in Chapter 6.4. The antiderivative of $1\/\left(u^2-1\right)$ is obtained from Table 3.10.1. The logarithmic form of the inverse hyperbolic tangent function is given in Table 2.10.4.

The substitution $v^2\=4\+9 x^2$ leads to an alternate solution not based on trig substitution. Since $2 v \mathrm{dv}\=18 x \mathrm{dx}$, the integral becomes

The conversion of $\frac{v^2}{v^2-4}$ to $1\+\frac{1}{v-2}-\frac{1}{v\+2}$ is again attributed to the algebraic technique of partial fraction decomposition.$$

Using the appropriate identity in Table 2.10.4, the alternate form of the solution, namely,

$\sqrt{4\+9 x^2}\+\ln \left(\sqrt{4\+9 x^2}-2\right)-\ln \left(\sqrt{4\+9 x^2}\+2\right)$

can be obtained from the Maple solution.

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

The stepwise solution provided by the  tutor when the Constant, Constant Multiple, and Sum rules are taken as Understood Rules begins with the substitution $u^2\=9 x^2\+4$ that results in the integral $\int \frac{u^2}{u^2-4} \mathrm{du}\=\int \left(1\+\frac{4}{u^2-4}\right) \mathrm{du}\=\int \left(1\+\frac{1}{u-2}-\frac{1}{u-2}\right) \mathrm{du}$, the last form being the result of a partial fraction decomposition.

On the other hand, Table 6.3.13(a) shows the result when the Change rule $x\=\frac{2}{3}\tan \left(\mathrm{\θ}\right)$ is imposed on the tutor. The further application of the Change rule with $u\=\cos \left(\mathrm{\θ}\right)$ gives a rational function that yields to a partial fraction decomposition.

Rather than simply evaluating each of the three integrals in the last line of Table 6.3.13(a), the tutor evaluates each by making detailed changes of variables. Clearly, the outcome has to be

$-\ln \(\left|u\+1\right|\)$ $\+\ln \(\left|u-1\right|\)$ $\+2\/u$ 

with $u\=\cos \left(\mathrm{\θ}\right)\=2\/\sqrt{9 x^2\+4}$, as per Figure 6.3.2.