$\int x \sqrt{9 x^2-4}\mathit{\ⅆ}x$

$\= \int \left(\frac{2}{3}\sec \left(\mathrm{\θ}\right)\right)\left(2 \tan \left(\mathrm{\θ}\right)\right)\left(\frac{2}{3}\sec \left(\mathrm{\θ}\right)\tan \left(\mathrm{\θ}\right) d\mathrm{\θ}\right)$

$\=\frac{8}{9}\int {\sec }^2\left(\mathrm{\θ}\right){\tan }^2\left(\mathrm{\θ}\right) d\mathrm{\θ}$

$\{\begin{array}{c}u\=\tan \left(\mathrm{\θ}\right)\\ \mathrm{du}\={\sec }^2\left(\mathrm{\θ}\right)d\mathrm{\θ}\end{array}$

$\=\frac{8}{9}\int u^2 \mathrm{du}$

$\=\frac{8}{9}\frac{u^3}{3}$

$\=\frac{8}{27}{\tan }^3\left(\mathrm{\θ}\right)$

$\=\frac{8}{27}{\left(\frac{\sqrt{9 x^2-4}}{2}\right)}^3$

$\=\frac{9 x^2-4}{27}\sqrt{9 x^2-4}$ 

Evaluate the given integral

$\int x \sqrt{9 x^2-4}\mathit{\ⅆ}x$ = $\frac{1}{27}\left(3x-2\right)\left(3x\+2\right)\sqrt{9x^2-4}$$\stackrel{\text{simplify}}{\=}$$\frac{1}{27}{\left(9x^2-4\right)}^{3\/2}$$$

Initialization

$\mathrm{with}\left(\mathrm{IntegrationTools}\right)\:$

$Q≔\int x \sqrt{9 x^2-4}\mathit{\ⅆ}x\:$

Change variables as per Table 6.3.1

$q_1≔\mathrm{Change}\left(Q\,x\=\frac{2}{3}\sec \left(\mathrm{\θ}\right)\right)$

$\∫\frac{4}{9}{\sec \left(\mathrm{\θ}\right)}^2\sqrt{4{\sec \left(\mathrm{\θ}\right)}^2-4}\tan \left(\mathrm{\θ}\right)\ⅆ\mathrm{\θ}$

$q_2≔\mathrm{simplify}\left(q_1\right) assuming \mathrm{\θ}\colon\colon \mathrm{RealRange}\left(0\,\frac{\mathrm{\π}}{2}\right)$

$\frac{8}{9}\∫\frac{{\sin \left(\mathrm{\θ}\right)}^2}{{\cos \left(\mathrm{\θ}\right)}^4}\ⅆ\mathrm{\θ}$

$q_3≔\mathrm{value}\left(q_2\right)$

$\frac{8}{27}\frac{{\sin \left(\mathrm{\θ}\right)}^3}{{\cos \left(\mathrm{\θ}\right)}^3}$

$\mathrm{simplify}\left(\mathrm{eval}\left(q_3\,\mathrm{\θ}\=\mathrm{arcsec}\left(\frac{3}{2}x\right)\right)\right) assuming x\colon\colon \mathrm{RealRange}\left(0\,\mathrm{\π}\/2\right)$

$\frac{1}{27}{\left(9x^2-4\right)}^{3\/2}$

$\begin{array}{cccc}\multicolumn{4}{c}{\int x\sqrt{9x^2-4}\ⅆx}\\ & \text{=}& \frac{\int u^2\ⅆu}{9}\hfill & \left[\mathrm{change}\,9x^2-4=u^2\,u\right]\\ & \text{=}& \frac{u^3}{27}\hfill & \left[\mathrm{power}\right]\\ & \text{=}& \frac{{\left(9x^2-4\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{27}\hfill & \left[\mathrm{revert}\right]\end{array}$

Table 6.3.19(a)   The substitution $u^2\=9 x^2-4$ made by the Integration Methods tutor

$\begin{array}{cccc}\multicolumn{4}{c}{\int x\sqrt{9x^2-4}\ⅆx}\\ & \text{=}& \frac{4\int {\sec \left(\mathrm{\theta }\right)}^2\sqrt{4{\sec \left(\mathrm{\theta }\right)}^2-4}\tan \left(\mathrm{\theta }\right)\ⅆ\mathrm{\theta }}{9}\hfill & \left[\mathrm{change}\,x=\frac{2\sec \left(\mathrm{\theta }\right)}{3}\right]\\ & \text{=}& \frac{8\int u^2\ⅆu}{9}\hfill & \left[\mathrm{change}\,u=\tan \left(\mathrm{\theta }\right)\,u\right]\\ & \text{=}& \frac{8u^3}{27}\hfill & \left[\mathrm{power}\right]\\ & \text{=}& \frac{8{\tan \left(\mathrm{\theta }\right)}^3}{27}\hfill & \left[\mathrm{revert}\right]\\ & \text{=}& \frac{{\left(9x^2-4\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{27}\hfill & \left[\mathrm{revert}\right]\end{array}$

Table 6.3.19(b)   Integration Methods tutor after $x\=\frac{2}{3}\sec \left(\mathrm{\theta }\right)$ is imposed