求常数 $A$ 和 $B$ 的值，使得对于所有 $x$ 的值，都有
$11 - 4x = A(4 - 2x) + B.$ 由此，求 $\int \frac{11 - 4x}{\sqrt{x(4 - x)}} \, \mathrm{d}x$。 使用代换 $x = \sin \theta$ 求 $\int \sqrt{1 - x^2} \, \mathrm{d}x$。 计算 $\int_{1}^{2} x \mathrm{e}^{2x} \, \mathrm{d}x$，结果用精确形式表示。

1. State the values of constants $A$ and $B$.

By expanding and comparing coefficients of $x$:
$-4 = -2A \implies A = 2$
By comparing constant terms:
$11 = 4A + B \implies 11 = 4(2) + B \implies B = 3$
Thus, $A = 2$ and $B = 3$.

2. Rewrite the given integral using the constants $A$ and $B$ and split it into two manageable integrals.

Using the identity, substitute $11 - 4x$ into the numerator:
$\int \frac{11 - 4x}{\sqrt{x(4 - x)}} \, \mathrm{d}x = \int \frac{2(4 - 2x) + 3}{\sqrt{4x - x^2}} \, \mathrm{d}x$
$= 2 \int (4 - 2x)(4x - x^2)^{-\frac{1}{2}} \, \mathrm{d}x + 3 \int \frac{1}{\sqrt{4x - x^2}} \, \mathrm{d}x$

3. Evaluate the first integral using the reverse chain rule.

The derivative of $4x - x^2$ is $4 - 2x$, so this integral is evaluated using $\int \mathrm{f}'(x)[\mathrm{f}(x)]^n \, \mathrm{d}x$:
$2 \int (4 - 2x)(4x - x^2)^{-\frac{1}{2}} \, \mathrm{d}x = 2 \left[ \frac{(4x - x^2)^{\frac{1}{2}}}{\frac{1}{2}} \right] = 4\sqrt{4x - x^2}$

4. Complete the square for the expression inside the square root of the second integral.

$4x - x^2 = -(x^2 - 4x)$
$= -\left[ (x - 2)^2 - 4 \right]$
$= 4 - (x - 2)^2 = 2^2 - (x - 2)^2$

5. Evaluate the second integral using a standard formula and write the overall answer.

Using the standard result $\int \frac{1}{\sqrt{a^2 - x^2}} \, \mathrm{d}x = \sin^{-1}\left(\frac{x}{a}\right)$:
$3 \int \frac{1}{\sqrt{2^2 - (x - 2)^2}} \, \mathrm{d}x = 3\sin^{-1}\left(\frac{x - 2}{2}\right)$
Combining the results, the final integral is:
$4\sqrt{4x - x^2} + 3\sin^{-1}\left(\frac{x - 2}{2}\right) + c$
where $c$ is an arbitrary constant.

1. Find the differential $\mathrm{d}x$ in terms of $\theta$.

$x = \sin \theta \implies \frac{\mathrm{d}x}{\mathrm{d}\theta} = \cos \theta$
$\mathrm{d}x = \cos \theta \, \mathrm{d}\theta$

2. Substitute $x$ and $\mathrm{d}x$ into the integral and simplify using the Pythagorean identity.

$\int \sqrt{1 - x^2} \, \mathrm{d}x = \int \sqrt{1 - \sin^2 \theta} (\cos \theta) \, \mathrm{d}\theta$
$= \int \sqrt{\cos^2 \theta} (\cos \theta) \, \mathrm{d}\theta$
$= \int \cos^2 \theta \, \mathrm{d}\theta$

3. Integrate with respect to $\theta$ using the double-angle identity for cosine.

$= \int \frac{1}{2}(1 + \cos 2\theta) \, \mathrm{d}\theta$
$= \frac{1}{2} \left( \theta + \frac{1}{2} \sin 2\theta \right) + C$

4. Express the result back in terms of $x$.

Using the double-angle identity for sine, $\sin 2\theta = 2 \sin \theta \cos \theta$:
$= \frac{1}{2} \theta + \frac{1}{4} (2 \sin \theta \cos \theta) + C$
$= \frac{1}{2} \theta + \frac{1}{2} \sin \theta \cos \theta + C$

Since $x = \sin \theta$, we have $\theta = \sin^{-1} x$ and $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - x^2}$.
$= \frac{1}{2} \sin^{-1} x + \frac{1}{2} x \sqrt{1 - x^2} + C$

1. Apply the integration by parts formula to the definite integral.

Let $u = x \implies \frac{\mathrm{d}u}{\mathrm{d}x} = 1$
Let $\frac{\mathrm{d}v}{\mathrm{d}x} = \mathrm{e}^{2x} \implies v = \frac{1}{2}\mathrm{e}^{2x}$

Using $\int u \frac{\mathrm{d}v}{\mathrm{d}x} \, \mathrm{d}x = uv - \int v \frac{\mathrm{d}u}{\mathrm{d}x} \, \mathrm{d}x$:
$\int_{1}^{2} x \mathrm{e}^{2x} \, \mathrm{d}x = \left[ \frac{1}{2} x \mathrm{e}^{2x} \right]_{1}^{2} - \int_{1}^{2} \frac{1}{2} \mathrm{e}^{2x} \, \mathrm{d}x$

2. Integrate the remaining exponential term.

$= \left[ \frac{1}{2} x \mathrm{e}^{2x} \right]_{1}^{2} - \left[ \frac{1}{4} \mathrm{e}^{2x} \right]_{1}^{2}$

3. Evaluate the definite integral by substituting the upper and lower limits to find the exact value.

Substituting $x = 2$ into the expression:
Upper limit evaluation $= \left( \frac{1}{2}(2)\mathrm{e}^4 - \frac{1}{4}\mathrm{e}^4 \right) = \mathrm{e}^4 - \frac{1}{4}\mathrm{e}^4 = \frac{3}{4}\mathrm{e}^4$

Substituting $x = 1$ into the expression:
Lower limit evaluation $= \left( \frac{1}{2}(1)\mathrm{e}^2 - \frac{1}{4}\mathrm{e}^2 \right) = \frac{1}{4}\mathrm{e}^2$

Subtracting the lower limit from the upper limit:
$\int_{1}^{2} x \mathrm{e}^{2x} \, \mathrm{d}x = \frac{3}{4}\mathrm{e}^4 - \frac{1}{4}\mathrm{e}^2$