求 $\sqrt{x^2 - 1}$ 关于 $x$ 的导数。 由此求 $\displaystyle \int x\cos^{-1}\left(\dfrac{1}{x}\right)\,dx$。 使用代换 $u = \dfrac{1}{x}$ 求 $\displaystyle \int_{\tfrac{1}{3}}^{6} \dfrac{1}{x\sqrt{x^2 - 9}}\,dx$ 的精确值。

1. Rewrite the expression with a fractional power and apply the chain rule

Let $y = \sqrt{x^2 - 1} = (x^2 - 1)^{\frac{1}{2}}$.

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}(x^2 - 1)^{-\frac{1}{2}} \cdot \frac{\mathrm{d}}{\mathrm{d}x}(x^2 - 1)$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}(x^2 - 1)^{-\frac{1}{2}} \cdot (2x)$

2. Simplify the resulting expression

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2x}{2\sqrt{x^2 - 1}} = \frac{x}{\sqrt{x^2 - 1}}$

1. Find the derivative of $\sqrt{x^2 - 1}$ from part (a)(i) to use later.

From part (a)(i), we calculate the derivative using the chain rule:
$\frac{\mathrm{d}}{\mathrm{d}x} \left(\sqrt{x^2-1}\right) = \frac{1}{2}(x^2-1)^{-1/2}(2x) = \frac{x}{\sqrt{x^2-1}}$

2. Set up integration by parts for the target integral.

We use integration by parts for $\int x \cos^{-1}\left(\frac{1}{x}\right) \,\mathrm{d}x$. Let $u = \cos^{-1}\left(\frac{1}{x}\right)$ and $\frac{\mathrm{d}v}{\mathrm{d}x} = x$.
Then $v = \frac{x^2}{2}$.
We differentiate $u$ using the chain rule (assuming $x > 0$ based on standard conventions for this type of problem):
$\frac{\mathrm{d}u}{\mathrm{d}x} = -\frac{1}{\sqrt{1 - \left(\frac{1}{x}\right)^2}} \left(-\frac{1}{x^2}\right) = \frac{1}{x^2 \sqrt{\frac{x^2-1}{x^2}}} = \frac{1}{x\sqrt{x^2-1}}$

3. Apply the integration by parts formula.

Substitute $u$, $v$, $\frac{\mathrm{d}u}{\mathrm{d}x}$ and $\frac{\mathrm{d}v}{\mathrm{d}x}$ into the integration by parts formula $\int u \frac{\mathrm{d}v}{\mathrm{d}x} \,\mathrm{d}x = uv - \int v \frac{\mathrm{d}u}{\mathrm{d}x} \,\mathrm{d}x$:
$\int x \cos^{-1}\left(\frac{1}{x}\right) \,\mathrm{d}x = \frac{x^2}{2}\cos^{-1}\left(\frac{1}{x}\right) - \int \frac{x^2}{2} \left( \frac{1}{x\sqrt{x^2-1}} \right) \,\mathrm{d}x$
$= \frac{x^2}{2}\cos^{-1}\left(\frac{1}{x}\right) - \frac{1}{2} \int \frac{x}{\sqrt{x^2-1}} \,\mathrm{d}x$

4. Use the result from part (a)(i) to evaluate the remaining integral.

Recognising the remaining integral is exactly the reverse process of part (a)(i), we have $\int \frac{x}{\sqrt{x^2-1}} \,\mathrm{d}x = \sqrt{x^2-1}$.
Substituting this back yields the final answer:
$\int x \cos^{-1}\left(\frac{1}{x}\right) \,\mathrm{d}x = \frac{x^2}{2}\cos^{-1}\left(\frac{1}{x}\right) - \frac{1}{2}\sqrt{x^2-1} + C$
where $C$ is an arbitrary constant of integration.