Let
\(I=\int \frac{x \log x}{\left(\sqrt{x^2-1}\right)^3} d x\)
Let \(\sqrt{x^2-1}=t \Rightarrow \frac{2 x d x}{2 \sqrt{x^2-1}}=d t \Rightarrow \frac{x d x}{\sqrt{x^2-1}}=d t\)
\(\begin{aligned}
& \therefore \quad I=\int \frac{\log x d t}{t^2} \\
&=\int \frac{\log \left(\sqrt{1+t^2}\right)}{t^2} d t=\int \log \sqrt{\left(1+t^2\right)} \cdot\left(\frac{1}{t^2}\right) d t \\
& {\left.\left[\because \sqrt{x^2-1}=t\right] \Rightarrow x^2=t^2+1 \Rightarrow x=\sqrt{1+t^2}\right] } \\
&=-\frac{1}{t} \log \sqrt{1+t^2}-\int \frac{2 t}{2 \sqrt{1+t^2}} \cdot\left(-\frac{1}{t}\right) \cdot \frac{1}{\sqrt{1+t^2}} d t \\
&=-\frac{1}{t} \log \sqrt{1+t^2}+\int \frac{d t}{1+t^2} \\
&=-\frac{1}{t} \log \sqrt{1+t^2}+\tan ^{-1}(t)+c \\
& \tan ^{-1}\left(\sqrt{x^2-1}\right)-\frac{\log x}{\sqrt{x^2-1}}+c \\
&=\sec ^{-1}(x)-\frac{\log x}{\sqrt{x^2-1}}+c
\end{aligned}\)