$\int \frac{\sqrt{9 x^2-4}}{x^2}\mathit{\ⅆ}x$

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

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

$\=3\int \frac{{\sec }^2\left(\mathrm{\θ}\right)-1}{\sec \left(\mathrm{\θ}\right)} d\mathrm{\θ}$

$\=3\int \left(\sec \left(\mathrm{\θ}\right)-\cos \left(\mathrm{\θ}\right)\right) d\mathrm{\θ}$

$\=3 \ln \left(\left|\sec \left(\mathrm{\θ}\right)\+\tan \left(\mathrm{\θ}\right)\right|\right)-3 \sin \left(\mathrm{\θ}\right)$ 

$\=3 \ln \left(\left|\frac{3}{2}x\+\frac{\sqrt{9 x^2-4}}{2}\right|\right)-3\left(\frac{\sqrt{9 x^2-4}}{3 x}\right)$

$\=3 \ln \left(\left|\frac{3}{2}x\+\frac{\sqrt{9 x^2-4}}{2}\right|\right)-\frac{\sqrt{9 x^2-4}}{x}$

Evaluate the given integral

$\int \frac{\sqrt{9 x^2-4}}{x^2}\mathit{\ⅆ}x$

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

$\stackrel{\text{simplify}}{\=}$

$\frac{3\ln \left(3x\+\sqrt{9x^2-4}\right)x-\sqrt{9x^2-4}}{x}$

$\stackrel{\text{expand}}{\=}$

$3\ln \left(3x\+\sqrt{9x^2-4}\right)-\frac{\sqrt{9x^2-4}}{x}$

Initialization

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

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

Change variables as per Table 6.3.1

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

$\∫\frac{3}{2}\frac{\sqrt{4{\sec \left(\mathrm{\θ}\right)}^2-4}\tan \left(\mathrm{\θ}\right)}{\sec \left(\mathrm{\θ}\right)}\ⅆ\mathrm{\θ}$

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

$3\left(\∫\frac{{\sin \left(\mathrm{\θ}\right)}^2}{\cos \left(\mathrm{\θ}\right)}\ⅆ\mathrm{\θ}\right)$

$q_3≔\mathrm{value}\left(q_2\right)$

$-3\sin \left(\mathrm{\θ}\right)\+3\ln \left(\sec \left(\mathrm{\θ}\right)\+\tan \left(\mathrm{\θ}\right)\right)$

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

$\frac{-3\ln \left(2\right)x\+3\ln \left(3x\+\sqrt{9x^2-4}\right)x-\sqrt{9x^2-4}}{x}$

$\stackrel{\text{expand}}{\=}$

$-3\ln \left(2\right)\+3\ln \left(3x\+\sqrt{9x^2-4}\right)-\frac{\sqrt{9x^2-4}}{x}$

$\begin{array}{c}\int \frac{\sqrt{9x^2-4}}{x^2}\mathit{\ⅆ}x\hfill \\ & \text{=}& \int \frac{-3u^4+24u^2-48}{u^5+8u^3+16u}\mathit{\ⅆ}u\hfill & \left[\mathrm{change}\,u=\sqrt{9x^2-4}-3x\,u\right]\\ & \text{=}& -3\int \frac{1}{u}\mathit{\ⅆ}u+48\int \frac{u}{{\left(u^2+4\right)}^2}\mathit{\ⅆ}u\hfill & \left[\mathrm{partialfractions}\right]\\ & \text{=}& -3\ln \left(u\right)+48\int \frac{u}{{\left(u^2+4\right)}^2}\mathit{\ⅆ}u\hfill & \left[\mathrm{power}\right]\\ & \text{=}& -3\ln \left(u\right)+24\int \frac{1}{v^2}\mathit{ }\mathrm{dv}\hfill & \left[\mathrm{change}\,v=u^2+4\,v\right]\\ & \text{=}& -3\ln \left(u\right)-\frac{24}{v}\hfill & \left[\mathrm{power}\right]\\ & \text{=}& -3\ln \left(u\right)-\frac{48}{2u^2+8}\hfill & \left[\mathrm{revert}\right]\end{array}$

Table 6.3.24(a)   The substitution $u\=\sqrt{9 x^2-4}-3 x$ made by the Integration Methods tutor

$\begin{array}{c}\int \frac{\sqrt{9x^2-4}}{x^2}\mathit{\ⅆ}x\hfill \\ & \text{=}& 3\int \frac{{\sin \left(\mathrm{\theta }\right)}^2}{\cos \left(\mathrm{\theta }\right)}\mathit{\ⅆ}\mathrm{\theta }\hfill & \left[\mathrm{change}\,x=\frac{2\sec \left(\mathrm{\theta }\right)}{3}\right]\\ & \text{=}& 3\int \frac{1}{\cos \left(\mathrm{\theta }\right)}\mathit{\ⅆ}\mathrm{\theta }-3\int \cos \left(\mathrm{\theta }\right)\mathit{\ⅆ}\mathrm{\theta }\hfill & \left[\mathrm{rewrite}\,\frac{{\sin \left(\mathrm{\theta }\right)}^2}{\cos \left(\mathrm{\theta }\right)}=\frac{1}{\cos \left(\mathrm{\theta }\right)}-\cos \left(\mathrm{\theta }\right)\right]\\ & \text{=}& 3\int \sec \left(\mathrm{\theta }\right)\mathit{\ⅆ}\mathrm{\theta }-3\int \cos \left(\mathrm{\theta }\right)\mathit{\ⅆ}\mathrm{\theta }\hfill & \left[\mathrm{rewrite}\,\frac{1}{\cos \left(\mathrm{\theta }\right)}=\sec \left(\mathrm{\theta }\right)\right]\end{array}$

Table 6.3.24(b)   Integration Methods tutor after $x\=\frac{2}{3}\sec \left(\mathrm{\theta }\right)$ is imposed