First, split the fraction:

\(\frac{x(x+1)}{x^2+4} = 1 + \frac{x-4}{x^2+4}\)

Integrate the first part:

\(\int 1 \, dx = x\)

For the second part, use partial fraction decomposition:

\(\int \frac{x-4}{x^2+4} \, dx = \frac{1}{2} \ln(x^2+4) - 2 \arctan\left(\frac{x}{2}\right)\)

Evaluate from 0 to 6:

\(\left[ x + \frac{1}{2} \ln(x^2+4) - 2 \arctan\left(\frac{x}{2}\right) \right]_{0}^{6}\)

Calculate at the bounds:

At \(x = 6\):

\(6 + \frac{1}{2} \ln(36+4) - 2 \arctan 3\)

At \(x = 0\):

\(0 + \frac{1}{2} \ln 4 - 2 \arctan 0\)

Subtract the results:

\(\left( 6 + \frac{1}{2} \ln 40 - 2 \arctan 3 \right) - \left( \frac{1}{2} \ln 4 \right)\)

Simplify:

\(6 + \frac{1}{2} \ln 10 - 2 \arctan 3\)

(i) The expansion of \(\cos(3x + x)\) is \(\cos 3x \cos x - \sin 3x \sin x\).

The expansion of \(\cos(3x - x)\) is \(\cos 3x \cos x + \sin 3x \sin x\).

Adding these, we get:

\(\cos(3x + x) + \cos(3x - x) = 2 \cos 3x \cos x\).

Thus, \(\frac{1}{2}(\cos 4x + \cos 2x) = \cos 3x \cos x\).

(ii) We have \(\int \cos 3x \cos x \, dx = \frac{1}{2} \int (\cos 4x + \cos 2x) \, dx\).

Integrating, we get:

\(\frac{1}{2} \left( \frac{1}{4} \sin 4x + \frac{1}{2} \sin 2x \right) = \frac{1}{8} \sin 4x + \frac{1}{4} \sin 2x\).

Substituting the limits \(-\frac{1}{6}\pi\) to \(\frac{1}{6}\pi\), we find:

\(\left[ \frac{1}{8} \sin 4x + \frac{1}{4} \sin 2x \right]_{-\frac{1}{6}\pi}^{\frac{1}{6}\pi} = \frac{3}{8}\sqrt{3}\).

(i) Start with the given equation: \(x = \ln(1-y) - \ln y\).

This can be rewritten using properties of logarithms: \(x = \ln \left( \frac{1-y}{y} \right)\).

Exponentiating both sides gives: \(e^x = \frac{1-y}{y}\).

Rearranging for \(y\), we get: \(y = \frac{1}{1 + e^x}\).

Substitute \(e^x = \frac{1}{e^{-x}}\) to obtain: \(y = \frac{e^{-x}}{1 + e^{-x}}\).

(ii) We need to evaluate \(\int_0^1 y \, dx\).

Using the expression for \(y\), we have \(y = \frac{e^{-x}}{1 + e^{-x}}\).

Let \(u = 1 + e^{-x}\), then \(du = -e^{-x} \, dx\), so \(dx = -\frac{du}{e^{-x}}\).

Substitute into the integral: \(\int y \, dx = \int \frac{e^{-x}}{u} \left(-\frac{du}{e^{-x}}\right) = -\int \frac{1}{u} \, du\).

This evaluates to \(-\ln |u| + C = -\ln(1 + e^{-x}) + C\).

Evaluate from 0 to 1:

At \(x = 1\), \(-\ln(1 + e^{-1})\).

At \(x = 0\), \(-\ln(1 + e^0) = -\ln(2)\).

Thus, \(\int_0^1 y \, dx = -\ln(1 + e^{-1}) + \ln(2)\).

Simplifying gives: \(\ln \left( \frac{2}{1 + e^{-1}} \right) = \ln \left( \frac{2e}{e+1} \right)\).

(i) Start with the left-hand side: \(\tan 2\theta - \tan \theta\).

Using the identity \(\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}\), substitute:

\(\frac{2\tan \theta}{1 - \tan^2 \theta} - \tan \theta\).

Combine into a single fraction:

\(\frac{2\tan \theta - \tan \theta (1 - \tan^2 \theta)}{1 - \tan^2 \theta} = \frac{2\tan \theta - \tan \theta + \tan^3 \theta}{1 - \tan^2 \theta}\).

Simplify the numerator:

\(\frac{\tan \theta (1 + \tan^2 \theta)}{1 - \tan^2 \theta}\).

Recognize that \(\sec 2\theta = \frac{1}{\cos 2\theta} = \frac{1}{1 - \tan^2 \theta}\).

Thus, \(\tan \theta \sec 2\theta = \frac{\tan \theta}{1 - \tan^2 \theta}\).

Therefore, \(\tan 2\theta - \tan \theta = \tan \theta \sec 2\theta\).

(ii) Consider the integral \(\int_{0}^{\frac{1}{6}\pi} \tan \theta \sec 2\theta \, d\theta\).

Substitute \(\sec 2\theta = \frac{1}{\cos 2\theta}\), so the integral becomes:

\(\int_{0}^{\frac{1}{6}\pi} \frac{\tan \theta}{\cos 2\theta} \, d\theta\).

Use the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and \(\cos 2\theta = 1 - 2\sin^2 \theta\).

Integrate to obtain:

\(-\frac{1}{2} \ln(\cos 2\theta) + \ln(\cos \theta)\).

Evaluate from \(0\) to \(\frac{1}{6}\pi\):

\(-\frac{1}{2} \ln(\cos(\frac{1}{3}\pi)) + \ln(\cos(\frac{1}{6}\pi))\).

Substitute the values: \(\cos(\frac{1}{3}\pi) = \frac{1}{2}\) and \(\cos(\frac{1}{6}\pi) = \frac{\sqrt{3}}{2}\).

Resulting in:

\(-\frac{1}{2} \ln(\frac{1}{2}) + \ln(\frac{\sqrt{3}}{2}) = \frac{1}{2} \ln \frac{3}{2}\).

(a) Use the identity \(\tan^2 2x = \sec^2 2x - 1\) to rewrite the integral as \(\int (4 + \sec^2 2x - 1) \, dx = \int (3 + \sec^2 2x) \, dx\).

The integral of \(3 \, dx\) is \(3x\), and the integral of \(\sec^2 2x \, dx\) is \(\frac{1}{2} \tan 2x\) (using substitution \(u = 2x\), \(du = 2 \, dx\)).

Thus, the integral is \(3x + \frac{1}{2} \tan 2x\).

(b) Use the identity \(\sin x \cos \frac{1}{2}\pi + \cos x \sin \frac{1}{6}\pi\) to simplify the integrand to \(\cos \frac{1}{6}\pi + \frac{\cos x \sin \frac{1}{6}\pi}{\sin x}\).

The integral becomes \(\int_{\frac{1}{4}\pi}^{\frac{1}{2}\pi} \left( \cos \frac{1}{6}\pi + \frac{\cos x \sin \frac{1}{6}\pi}{\sin x} \right) \, dx\).

Integrate to obtain \(\cos \frac{1}{6}\pi x + \sin \frac{1}{6}\pi \ln(\sin x)\).

Apply the limits and simplify to obtain \(\frac{1}{8} \pi \sqrt{3} - \frac{1}{2} \ln \left( \frac{1}{\sqrt{2}} \right)\).