1. Factorise the denominator to match the standard integration form $\frac{1}{a^2 - x^2}$.

$\int \frac{1}{9 - 4x^2} \text{ d}x = \frac{1}{4} \int \frac{1}{\frac{9}{4} - x^2} \text{ d}x$

2. Apply the standard formula $\int \frac{1}{a^2 - x^2} \text{ d}x = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| + C$ with $a = \frac{3}{2}$.

$= \frac{1}{4} \left( \frac{1}{2\left(\frac{3}{2}\right)} \ln \left| \frac{\frac{3}{2} + x}{\frac{3}{2} - x} \right| \right) + C$

3. Simplify the coefficient and the fraction within the logarithm.

1. Find the differential $dx$ in terms of $\theta$.

Given $\sqrt{x} = \cos\theta$, squaring both sides gives $x = \cos^2\theta$.
Differentiating with respect to $\theta$:
$\frac{\text{d}x}{\text{d}\theta} = 2\cos\theta(-\sin\theta) = -2\cos\theta\sin\theta$
$\text{d}x = -2\cos\theta\sin\theta \text{ d}\theta$

2. Substitute into the integral and simplify.

$\int \sqrt{\frac{x}{1-x}} \text{ d}x = \int \sqrt{\frac{\cos^2\theta}{1-\cos^2\theta}} (-2\cos\theta\sin\theta) \text{ d}\theta$
Using the identity $1-\cos^2\theta = \sin^2\theta$:
$= \int \sqrt{\frac{\cos^2\theta}{\sin^2\theta}} (-2\cos\theta\sin\theta) \text{ d}\theta$
$= \int \frac{\cos\theta}{\sin\theta} (-2\cos\theta\sin\theta) \text{ d}\theta$
$= \int -2\cos^2\theta \text{ d}\theta$

3. Integrate using the double angle identity.

Using $\cos 2\theta = 2\cos^2\theta - 1$, we can rewrite $-2\cos^2\theta$ as $-(1 + \cos 2\theta)$:
$= \int -(1 + \cos 2\theta) \text{ d}\theta$
$= -\left(\theta + \frac{1}{2}\sin 2\theta\right) + C$
$= -\theta - \sin\theta\cos\theta + C$

4. Back-substitute to express the answer in terms of $x$.

From $\sqrt{x} = \cos\theta$, we have $\theta = \cos^{-1}(\sqrt{x})$.
We also need $\sin\theta$, which is $\sqrt{1-\cos^2\theta} = \sqrt{1-x}$.
Substituting these expressions back:
$-\theta - \sin\theta\cos\theta + C = -\cos^{-1}(\sqrt{x}) - \sqrt{1-x}(\sqrt{x}) + C$
$= -\cos^{-1}(\sqrt{x}) - \sqrt{x(1-x)} + C$

1. Apply integration by parts for the first time.

Let $u = (\ln x)^2 \implies \frac{\text{d}u}{\text{d}x} = \frac{2\ln x}{x}$
Let $\frac{\text{d}v}{\text{d}x} = x \implies v = \frac{1}{2}x^2$

$\int_{1}^{\text{e}} x(\ln x)^2 \text{ d}x = \left[ \frac{1}{2}x^2(\ln x)^2 \right]_{1}^{\text{e}} - \int_{1}^{\text{e}} \frac{1}{2}x^2 \left( \frac{2\ln x}{x} \right) \text{ d}x$
$= \left[ \frac{1}{2}x^2(\ln x)^2 \right]_{1}^{\text{e}} - \int_{1}^{\text{e}} x\ln x \text{ d}x$

2. Apply integration by parts a second time to evaluate $\int x \ln x \text{ d}x$.

Let $u = \ln x \implies \frac{\text{d}u}{\text{d}x} = \frac{1}{x}$
Let $\frac{\text{d}v}{\text{d}x} = x \implies v = \frac{1}{2}x^2$

$\int_{1}^{\text{e}} x\ln x \text{ d}x = \left[ \frac{1}{2}x^2\ln x \right]_{1}^{\text{e}} - \int_{1}^{\text{e}} \frac{1}{2}x^2 \left( \frac{1}{x} \right) \text{ d}x$
$= \left[ \frac{1}{2}x^2\ln x \right]_{1}^{\text{e}} - \int_{1}^{\text{e}} \frac{1}{2}x \text{ d}x$
$= \left[ \frac{1}{2}x^2\ln x - \frac{1}{4}x^2 \right]_{1}^{\text{e}}$

3. Combine the results and substitute the limits $1$ and $\text{e}$ to find the exact value.

$\int_{1}^{\text{e}} x(\ln x)^2 \text{ d}x = \left[ \frac{1}{2}x^2(\ln x)^2 - \left( \frac{1}{2}x^2\ln x - \frac{1}{4}x^2 \right) \right]_{1}^{\text{e}}$
$= \left[ \frac{1}{2}x^2(\ln x)^2 - \frac{1}{2}x^2\ln x + \frac{1}{4}x^2 \right]_{1}^{\text{e}}$

Substitute the upper limit $\text{e}$ and lower limit $1$:
$= \left( \frac{1}{2}\text{e}^2(1)^2 - \frac{1}{2}\text{e}^2(1) + \frac{1}{4}\text{e}^2 \right) - \left( 0 - 0 + \frac{1}{4}(1)^2 \right)$
$= \frac{1}{4}\text{e}^2 - \frac{1}{4}$
$= \frac{\text{e}^2 - 1}{4}$