To solve \(\int_{1}^{\infty} (4x + 2)^{-2} \, dx\), we first find the antiderivative of \((4x + 2)^{-2}\).

Let \(u = 4x + 2\), then \(du = 4 \, dx\) or \(dx = \frac{1}{4} \, du\).

The integral becomes \(\int (4x + 2)^{-2} \, dx = \int u^{-2} \cdot \frac{1}{4} \, du\).

This simplifies to \(\frac{1}{4} \int u^{-2} \, du\).

The antiderivative of \(u^{-2}\) is \(-u^{-1}\), so we have:

\(\frac{1}{4} \cdot (-u^{-1}) = -\frac{1}{4u}\).

Substitute back \(u = 4x + 2\):

\(-\frac{1}{4(4x + 2)} = -\frac{1}{16x + 8}\).

Now, evaluate the definite integral from 1 to \(\infty\):

\(\left[ -\frac{1}{16x + 8} \right]_{1}^{\infty} = \left( 0 - \left(-\frac{1}{24}\right) \right) = \frac{1}{24}\).

To integrate \(\int \left( 4x + \frac{6}{x^2} \right) \, dx\), we integrate each term separately.

1. Integrate \(4x\):

\(\int 4x \, dx = 4 \cdot \frac{x^2}{2} = 2x^2\).

2. Integrate \(\frac{6}{x^2}\):

Rewrite as \(6x^{-2}\).

\(\int 6x^{-2} \, dx = 6 \cdot \frac{x^{-1}}{-1} = -6x^{-1}\).

Combine the results:

\(2x^2 - 6x^{-1} + c\), where \(c\) is the constant of integration.

To solve \(\int_{1}^{\infty} \frac{1}{(3x - 2)^{\frac{3}{2}}} \, dx\), we first perform a substitution. Let \(u = 3x - 2\), then \(du = 3 \, dx\) or \(dx = \frac{du}{3}\).

The limits of integration change as follows: when \(x = 1\), \(u = 1\), and as \(x \to \infty\), \(u \to \infty\).

The integral becomes:

\(\int_{1}^{\infty} \frac{1}{u^{\frac{3}{2}}} \cdot \frac{1}{3} \, du = \frac{1}{3} \int_{1}^{\infty} u^{-\frac{3}{2}} \, du\)

Integrating \(u^{-\frac{3}{2}}\), we get:

\(\frac{1}{3} \left[ \frac{u^{-\frac{1}{2}}}{-\frac{1}{2}} \right]_{1}^{\infty} = \frac{1}{3} \left[ -2u^{-\frac{1}{2}} \right]_{1}^{\infty}\)

Evaluating the limits:

\(\frac{1}{3} \left[ -2(\infty)^{-\frac{1}{2}} + 2(1)^{-\frac{1}{2}} \right] = \frac{1}{3} \left[ 0 + 2 \right] = \frac{2}{3}\)

First, integrate the function \(\frac{1}{k} x^{\frac{1}{2}} + x^{-\frac{1}{2}} + \frac{1}{k^2}\):

\(\int \left( \frac{1}{k} x^{\frac{1}{2}} + x^{-\frac{1}{2}} + \frac{1}{k^2} \right) \, dx = \left[ \frac{2x^{\frac{3}{2}}}{3k} \right] + \left[ 2x^{\frac{1}{2}} \right] + \left[ \frac{x}{k^2} \right]\)

Apply the limits \(\frac{1}{4}k^2 \to k^2\) to the integrated expression:

\(\left( \frac{2k^2}{3} + 2k + 1 \right) - \left( \frac{k^2}{12} + k + \frac{1}{4} \right)\)

This simplifies to:

\(\frac{7}{12} k^2 + k + \frac{3}{4}\)

Equate this to \(\frac{13}{12}\) and simplify to a quadratic equation:

\(\frac{7}{12} k^2 + k + \frac{3}{4} = \frac{13}{12}\)

\(7k^2 + 12k - 4 = 0\)

Solving the quadratic equation \((7k - 2)(k + 2) = 0\), we find:

\(k = \frac{2}{7}\) only (or 0.286)

First, rewrite the integrand: \(\sqrt{x} = x^{1/2}\) and \(\frac{2}{\sqrt{x}} = 2x^{-1/2}\).

Integrate each term separately:

\(\int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}\).

\(\int 2x^{-1/2} \, dx = 2 \cdot \frac{x^{1/2}}{1/2} = 4x^{1/2}\).

Combine the integrals: \(\frac{2}{3}x^{3/2} + 4x^{1/2}\).

Evaluate from 1 to 4:

\(\left[ \frac{2}{3}x^{3/2} + 4x^{1/2} \right]_{1}^{4} = \left( \frac{2}{3}(4)^{3/2} + 4(4)^{1/2} \right) - \left( \frac{2}{3}(1)^{3/2} + 4(1)^{1/2} \right)\).

Calculate each part:

\((4)^{3/2} = 8\) and \((4)^{1/2} = 2\).

\(\frac{2}{3} \times 8 + 4 \times 2 = \frac{16}{3} + 8 = \frac{40}{3}\).

\(\frac{2}{3} \times 1 + 4 \times 1 = \frac{2}{3} + 4 = \frac{14}{3}\).

Subtract: \(\frac{40}{3} - \frac{14}{3} = \frac{26}{3}\).