求 $\displaystyle \int \dfrac{6+2x}{\sqrt{1-4x-x^2}}\,\mathrm{d}x$。 上图所示为由方程为 $y=\dfrac{\sqrt{\ln x}}{x}$（$x \ge 1$）的曲线 $C$、$x$ 轴以及方程为 $y=\dfrac{1}{e(e-2)}(x-2)$ 的直线 $L$ 所围成的区域 $R$。求当 $R$ 绕 $x$ 轴完全旋转一周时所形成的旋转体的准确体积。

1. Express the numerator in terms of the derivative of the quadratic expression inside the square root.

$\frac{\mathrm{d}}{\mathrm{d}x}(1-4x-x^2) = -4-2x$
$6+2x = -(-4-2x) + 2$
$\int \frac{6+2x}{\sqrt{1-4x-x^2}} \,\mathrm{d}x = \int \frac{-(-4-2x)}{\sqrt{1-4x-x^2}} \,\mathrm{d}x + \int \frac{2}{\sqrt{1-4x-x^2}} \,\mathrm{d}x$

2. Integrate the first term using recognition of the form $f'(x)[f(x)]^n$.

$-\int (-4-2x)(1-4x-x^2)^{-\frac{1}{2}} \,\mathrm{d}x = -\frac{(1-4x-x^2)^{\frac{1}{2}}}{\frac{1}{2}} = -2\sqrt{1-4x-x^2}$

3. Complete the square for the quadratic expression to prepare the second term for integration.

$1-4x-x^2 = 1 - (x^2+4x)$
$= 1 - \left((x+2)^2 - 4\right)$
$= 5 - (x+2)^2$

4. Integrate the second term using the standard formula for inverse sine.

$\int \frac{2}{\sqrt{5-(x+2)^2}} \,\mathrm{d}x = 2\sin^{-1}\left(\frac{x+2}{\sqrt{5}}\right)$

5. Combine the integrated parts and include the constant of integration.

$\int \frac{6+2x}{\sqrt{1-4x-x^2}} \,\mathrm{d}x = -2\sqrt{1-4x-x^2} + 2\sin^{-1}\left(\frac{x+2}{\sqrt{5}}\right) + C$

1. Set up the expression for the required volume of revolution.

The volume of the solid is the difference between the volume generated by rotating curve $C$ from $x=1$ to $x=e$, and the volume generated by rotating line $L$ from $x=2$ to $x=e$.
$V = \pi \int_{1}^{e} \left( \frac{\sqrt{\ln x}}{x} \right)^2 \, \mathrm{d}x - \pi \int_{2}^{e} \left( \frac{1}{e(e-2)}(x-2) \right)^2 \, \mathrm{d}x$

2. Integrate the function for the curve $C$ using integration by parts.

For the first integral, we need to evaluate $\int \frac{\ln x}{x^2} \, \mathrm{d}x$.
Let $u = \ln x \implies \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{x}$
Let $\frac{\mathrm{d}v}{\mathrm{d}x} = x^{-2} \implies v = -x^{-1} = -\frac{1}{x}$

Using the integration by parts formula $\int u \, \mathrm{d}v = uv - \int v \, \mathrm{d}u$:
$\int \frac{\ln x}{x^2} \, \mathrm{d}x = -\frac{\ln x}{x} - \int \left( -\frac{1}{x} \right) \left( \frac{1}{x} \right) \, \mathrm{d}x$
$= -\frac{\ln x}{x} + \int x^{-2} \, \mathrm{d}x = -\frac{\ln x}{x} - \frac{1}{x} = -\frac{\ln x + 1}{x}$

3. Evaluate the limits for the volume generated by $C$.

$V_C = \pi \left[ -\frac{\ln x + 1}{x} \right]_{1}^{e}$
$= \pi \left( -\frac{\ln e + 1}{e} - \left( -\frac{\ln 1 + 1}{1} \right) \right)$
$= \pi \left( -\frac{2}{e} + 1 \right) = \pi \left( 1 - \frac{2}{e} \right)$

4. Calculate the volume generated by rotating the line $L$.

The solid generated by rotating the straight line $L$ between $x=2$ and $x=e$ is a right circular cone.
The base radius at $x=e$ is $r = \frac{1}{e(e-2)}(e-2) = \frac{1}{e}$.
The height of the cone is $h = e - 2$.
$V_L = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left( \frac{1}{e} \right)^2 (e - 2) = \frac{\pi (e - 2)}{3e^2}$
(Note: Evaluating $\pi \int_{2}^{e} \frac{(x-2)^2}{e^2(e-2)^2} \, \mathrm{d}x$ yields the exact same expression).

5. Subtract the two volumes to find the exact total volume of revolution.

$V = V_C - V_L = \pi \left( 1 - \frac{2}{e} \right) - \frac{\pi(e - 2)}{3e^2}$
$= \pi \left( \frac{3e^2 - 6e}{3e^2} - \frac{e - 2}{3e^2} \right)$
$= \pi \left( \frac{3e^2 - 7e + 2}{3e^2} \right)$