求 $\int \frac{x}{x^2 + 4x + 7} \, \mathrm{d}x$。 利用分部积分法求 $\int_0^a x\sqrt{a-x} \, \mathrm{d}x$。

1. Split the integral into a logarithmic form and a fraction with a constant numerator.

The derivative of the denominator $x^2 + 4x + 7$ is $2x + 4$. We can rewrite the numerator to involve this derivative:
$\int \frac{x}{x^2 + 4x + 7} \, \mathrm{d}x = \int \frac{\frac{1}{2}(2x + 4) - 2}{x^2 + 4x + 7} \, \mathrm{d}x$
$= \frac{1}{2}\int \frac{2x + 4}{x^2 + 4x + 7} \, \mathrm{d}x - 2\int \frac{1}{x^2 + 4x + 7} \, \mathrm{d}x$

2. Integrate the first term and complete the square for the denominator of the second term.

For the first integral, we use $\int \frac{\mathrm{f}'(x)}{\mathrm{f}(x)} \, \mathrm{d}x = \ln|\mathrm{f}(x)| + C$:
$\frac{1}{2}\int \frac{2x + 4}{x^2 + 4x + 7} \, \mathrm{d}x = \frac{1}{2}\ln(x^2 + 4x + 7)$
(Note: $x^2+4x+7 > 0$ for all real $x$, so absolute value signs are optional.)

For the second integral, complete the square for the denominator:
$x^2 + 4x + 7 = (x + 2)^2 + 3$

3. Evaluate the second integral using the standard arctan formula and state the final result.

Evaluate the second integral using $\int \frac{1}{x^2 + a^2} \, \mathrm{d}x = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$:
$2\int \frac{1}{(x + 2)^2 + (\sqrt{3})^2} \, \mathrm{d}x = \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{x + 2}{\sqrt{3}}\right)$

Combining both parts and adding the constant of integration $C$:
$\int \frac{x}{x^2 + 4x + 7} \, \mathrm{d}x = \frac{1}{2}\ln(x^2 + 4x + 7) - \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{x + 2}{\sqrt{3}}\right) + C$

1. Choose parts $u$ and $\frac{\mathrm{d}v}{\mathrm{d}x}$ and find $\frac{\mathrm{d}u}{\mathrm{d}x}$ and $v$.

Let $u = x$ and $\frac{\mathrm{d}v}{\mathrm{d}x} = (a-x)^{\frac{1}{2}}$.
Then $\frac{\mathrm{d}u}{\mathrm{d}x} = 1$.
Integrating $\frac{\mathrm{d}v}{\mathrm{d}x}$ gives $v = \int (a-x)^{\frac{1}{2}} \, \mathrm{d}x = -\frac{2}{3}(a-x)^{\frac{3}{2}}$.

2. Apply the integration by parts formula $\int u \frac{\mathrm{d}v}{\mathrm{d}x} \, \mathrm{d}x = uv - \int v \frac{\mathrm{d}u}{\mathrm{d}x} \, \mathrm{d}x$.

$\int_0^a x\sqrt{a-x} \, \mathrm{d}x = \left[ -\frac{2}{3}x(a-x)^{\frac{3}{2}} \right]_0^a - \int_0^a -\frac{2}{3}(a-x)^{\frac{3}{2}} \, \mathrm{d}x$

3. Evaluate the boundary term and perform the remaining integration.

Evaluate the first term:
$\left[ -\frac{2}{3}x(a-x)^{\frac{3}{2}} \right]_0^a = \left( -\frac{2}{3}a(a-a)^{\frac{3}{2}} \right) - \left( -\frac{2}{3}(0)(a-0)^{\frac{3}{2}} \right) = 0 - 0 = 0$

Evaluate the second integral:
$\int_0^a \frac{2}{3}(a-x)^{\frac{3}{2}} \, \mathrm{d}x = \left[ \frac{2}{3} \cdot \frac{(a-x)^{\frac{5}{2}}}{\frac{5}{2}} \cdot (-1) \right]_0^a = \left[ -\frac{4}{15}(a-x)^{\frac{5}{2}} \right]_0^a$

4. Substitute the limits to calculate the final value.

Substitute $x = a$ and $x = 0$:
$-\frac{4}{15}(a-a)^{\frac{5}{2}} - \left( -\frac{4}{15}(a-0)^{\frac{5}{2}} \right)$
$= 0 + \frac{4}{15}a^{\frac{5}{2}}$
$= \frac{4}{15}a^{\frac{5}{2}}$