求下列积分。

1. Find the derivative of the denominator.

Let $f(x) = \sin 3x + \cot 3x$.
Then $f'(x) = 3\cos 3x - 3\text{cosec}^2 3x = 3(\cos 3x - \text{cosec}^2 3x)$.
Notice that the numerator is $\frac{1}{3}f'(x)$.

2. Evaluate the integral using the standard logarithmic integrand rule.

$\int \frac{\cos 3x - \text{cosec}^2 3x}{\sin 3x + \cot 3x} \text{ d}x$
$= \frac{1}{3} \int \frac{3(\cos 3x - \text{cosec}^2 3x)}{\sin 3x + \cot 3x} \text{ d}x$
$= \frac{1}{3} \ln |\sin 3x + \cot 3x| + C$

1. Split the fraction into two separate integrals.

$\int \frac{1-x}{\sqrt{1-16x^2}} \text{ d}x = \int \frac{1}{\sqrt{1-16x^2}} \text{ d}x - \int \frac{x}{\sqrt{1-16x^2}} \text{ d}x$

2. Evaluate the first integral using the standard formula for inverse sine.

$\int \frac{1}{\sqrt{1-16x^2}} \text{ d}x = \int \frac{1}{\sqrt{1-(4x)^2}} \text{ d}x = \frac{1}{4} \sin^{-1}(4x)$

3. Evaluate the second integral by recognising it as the derivative of the inner function (reverse chain rule).

4. Combine the evaluated integrals, ensuring correct signs, and add the constant of integration.

1. Set up integration by parts by choosing $u$ and $\text{d}v$.

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

Let $\frac{\text{d}v}{\text{d}x} = (1-x)^{-2} \implies v = \int (1-x)^{-2} \text{ d}x = \frac{(1-x)^{-1}}{-1 \cdot (-1)} = \frac{1}{1-x}$

2. Apply the integration by parts formula.

$\int u \frac{\text{d}v}{\text{d}x} \text{ d}x = uv - \int v \frac{\text{d}u}{\text{d}x} \text{ d}x$
$\int (1-x)^{-2} \ln x \text{ d}x = \frac{\ln x}{1-x} - \int \frac{1}{x(1-x)} \text{ d}x$

3. Use partial fractions to decompose the remaining integrand.

Let $\frac{1}{x(1-x)} = \frac{A}{x} + \frac{B}{1-x}$
$1 = A(1-x) + Bx$
Substitute $x = 0 \implies A = 1$
Substitute $x = 1 \implies B = 1$
$\frac{1}{x(1-x)} = \frac{1}{x} + \frac{1}{1-x}$

4. Integrate the partial fractions and simplify to obtain the final answer.

$\int (1-x)^{-2} \ln x \text{ d}x = \frac{\ln x}{1-x} - \int \left( \frac{1}{x} + \frac{1}{1-x} \right) \text{ d}x$
$= \frac{\ln x}{1-x} - (\ln x - \ln |1-x|) + C$
$= \frac{\ln x}{1-x} - \frac{(1-x)\ln x}{1-x} + \ln |1-x| + C$
$= \frac{x \ln x}{1-x} + \ln |1-x| + C$