Answer
$\displaystyle \frac{2x^{2}-1}{x(x^{2}-1)}$
Work Step by Step
We use the theorem: $\displaystyle \frac{d}{dx}[\ln u]=\frac{1}{u}\cdot\frac{du}{dx}=\frac{u^{\prime}}{u},\ \ \ (u>0)$
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Here, $u(x)=x\sqrt{x^{2}-1}= \sqrt{x^{2}(x^{2}-1)}=\sqrt{x^{4}-x^{2}}=(x^{4}-x^{2})^{1/2}$
$\displaystyle \frac{du}{dx}$=...chain rule...=$\displaystyle \frac{1}{2}(x^{4}-x^{2})^{-1/2}\cdot(4x^{3}-2x)$
$=\displaystyle \frac{4x^3-2x}{2\sqrt{x^{4}-x^{2}}}=\frac{2x(2x^{2}-1 )}{ 2x\sqrt{x^{2}-1}}=\frac{ 2x^{2}-1 }{\sqrt{x^{2}-1}}$
So,
$\displaystyle \frac{d}{dx}[\ln(x\sqrt{x^{2}-1})]=\frac{1}{x\sqrt{x^{2}-1}}\cdot\frac{ 2x^{2}-1 }{\sqrt{x^{2}-1}}=\frac{2x^{2}-1}{x(x^{2}-1)}$
