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This is so well explained
alternative solution (perhaps more long winded but a bit different nonetheless)
(i) we have the integral
\begin{align}
\int\dfrac{dx}{x^{3/2}\sqrt{x-1}}
\\ x=\dfrac{1}{1-u}\implies dx=\dfrac{du}{(1-u)^2}
\\\therefore \int\dfrac{dx}{x^{3/2}\sqrt{x-1}}
&=\int\dfrac{(1-u)^{-2}du}{\sqrt{\dfrac{u}{1-u}}(1-u)^{-3/2}}
\\&=\int\dfrac{(1-u)^{-2}(1-u)^{3/2}\sqrt{(1-u)}du}{\sqrt{u}}
\\&=\int\dfrac{1}{\sqrt{u}}du
\\&=2\sqrt{u}+c
\\&=2\sqrt{\dfrac{x-1}{x}}+c
\end{align}
(ii) we now have the integral
\begin{align}
\int\dfrac{dx}{(x-2)^{3/2}\sqrt{x+1}}
\\\gamma=x-2\implies d\gamma=dx
\\\therefore\int\dfrac{dx}{(x-2)^{3/2}\sqrt{x+1}}
=\int\dfrac{d\gamma}{\gamma^{3/2}\sqrt{\gamma+3}}
\\\alpha=\dfrac{1}{\gamma + 3} \implies d\alpha=-\dfrac{d\gamma}{(\gamma+3)^{2}} \iff d\gamma=-(\gamma+3)^{2} d\alpha=-\alpha^{-2} d\alpha
\\\therefore\int\dfrac{dx}{\gamma^{3/2}\sqrt{\gamma+3}}
&=\int\dfrac{-\alpha^{-2}d\alpha}{\left(\frac{1-3\alpha}{\alpha}\right)^{3/2}\alpha^{-1/2}}
\\&=\int\dfrac{-\alpha^{2}d\alpha}{(1-3\alpha)^{3/2}\alpha^{2}}
\\&=\int -(1-3\alpha)^{-3/2}d\alpha
\\&=\dfrac{-2}{3} (1-3\alpha)^{-1/2}+c
\\&=\dfrac{-2}{3} \sqrt{\dfrac{x+1}{x-2}}+c
\end{align}
(iii) finally we have
\begin{align}
\int_{2}^{\infty}\dfrac{dx}{(x-1)\sqrt{x-2}\sqrt{3x-2}}
\\u=x-1 \implies du=dx
\\\therefore \int_{2}^{\infty} \dfrac{dx}{(x-1)\sqrt{x-2}\sqrt{3x-2}}
=\int_{1}^{\infty}\dfrac{du}{u \sqrt{u-1}\sqrt{3u+1}}
\\w=\dfrac{1}{3u+1}\implies du=\dfrac{-1}{3} w^{-2}dw
\\\therefore \int_{1/4}^{0}\dfrac{-\frac{1}{3}w^{-2}dw}{\left(\dfrac{1-w}{3w}\right)\sqrt{\dfrac{1-4w}{3w}}\left(\dfrac{1}{\sqrt{w}}\right)}
&=\dfrac{1}{3}\int_{0}^{1/4}\dfrac{w^{-2}w^{2}(3)^{3/2}dw}{(1-w)\sqrt{1-4w}}
\\&=\sqrt{3}\int_{0}^{1/4}\dfrac{dw}{(1-w)\sqrt{1-4w}}
\\\beta=\dfrac{1}{1-4w} \implies dw=\dfrac{1}{4}\beta^{-2}
\\\therefore \sqrt{3}\int_{0}^{1/4}\dfrac{dw}{(1-w)\sqrt{1-4w}}
&=\dfrac{\sqrt{3}}{4} \int_{1}^{\infty}\dfrac{\beta^{-2}d\beta}{\left(\dfrac{3\beta+1}{4\beta}\right)\beta^{-1/2}}
\\&=\dfrac{\sqrt{3}}{4} \int_{1}^{\infty}\dfrac{\beta^{-2}(4 {\beta})\sqrt{\beta}d \beta}{3 \beta +1}
\\&=\sqrt{3}\int_{1}^{\infty}\dfrac{d\beta}{\sqrt{\beta}(3\beta+1)}
\\\beta=\dfrac{1}{3}\tan^2\theta \implies d\beta=\dfrac{2}{3}\tan \theta \sec^{2}\theta d\theta
\\\therefore \sqrt{3}\int_{1}^{\infty}\dfrac{d\beta}{\sqrt{\beta}(3\beta+1)}
&=\sqrt{3}\int_{\pi/3}^{\pi/2}\dfrac{\frac{2}{3}\tan\theta \sec^{2}\theta d\theta}{\frac{1}{\sqrt{3}} \tan \theta (\tan^{2} \theta +1)}
\\&=2\int_{\pi/3}^{\pi/2}\dfrac{\tan\theta \sec^{2}\theta d\theta}{\tan \theta \sec^{2} \theta}
\\&=2\int_{\pi/3}^{\pi/2} 1 d \theta
\\&=2(\pi/2-\pi/3)
\\&=\dfrac{2\pi}{6}
\\&=\dfrac{\pi}{3}
\end{align}
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