To find \(\int \frac{2}{\sqrt{5x - 6}} \, dx\), we use substitution. Let \(u = 5x - 6\), then \(du = 5 \, dx\) or \(dx = \frac{du}{5}\).

The integral becomes \(\int \frac{2}{\sqrt{u}} \cdot \frac{1}{5} \, du = \frac{2}{5} \int u^{-1/2} \, du\).

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

\(\frac{2}{5} \cdot 2u^{1/2} = \frac{4}{5} \sqrt{u}\).

Substituting back for \(u\), we get \(\frac{4}{5} \sqrt{5x - 6}\).

Now, evaluate \(\int_{2}^{3} \frac{2}{\sqrt{5x - 6}} \, dx\):

\(\left[ \frac{4}{5} \sqrt{5x - 6} \right]_{2}^{3} = \frac{4}{5} \left( \sqrt{9} - \sqrt{4} \right)\).

\(= \frac{4}{5} (3 - 2) = \frac{4}{5} \cdot 1 = \frac{4}{5} = 0.8\).

To find \(\int (3x - 2)^5 \, dx\), use the substitution method. Let \(u = 3x - 2\), then \(\frac{du}{dx} = 3\) or \(dx = \frac{du}{3}\).

Thus, \(\int (3x - 2)^5 \, dx = \int u^5 \cdot \frac{du}{3} = \frac{1}{3} \int u^5 \, du\).

The integral of \(u^5\) is \(\frac{u^6}{6}\), so:

\(\frac{1}{3} \cdot \frac{u^6}{6} = \frac{(3x - 2)^6}{18} + C\).

Now, evaluate \(\int_0^1 (3x - 2)^5 \, dx\):

\(\left[ \frac{(3x - 2)^6}{18} \right]_0^1 = \frac{(3 \cdot 1 - 2)^6}{18} - \frac{(3 \cdot 0 - 2)^6}{18}\).

Calculate the values:

At \(x = 1\), \((3 \cdot 1 - 2)^6 = 1^6 = 1\).

At \(x = 0\), \((3 \cdot 0 - 2)^6 = (-2)^6 = 64\).

So, \(\frac{1}{18} - \frac{64}{18} = \frac{1 - 64}{18} = \frac{-63}{18} = -\frac{7}{2}\).

Therefore, the value is \(-\frac{3}{2}\).

To solve \(\int \left( x^3 + \frac{1}{x^3} \right) \, dx\), we integrate each term separately.

First, integrate \(x^3\):

\(\int x^3 \, dx = \frac{x^4}{4} + C_1\).

Next, integrate \(\frac{1}{x^3} = x^{-3}\):

\(\int x^{-3} \, dx = \frac{x^{-2}}{-2} + C_2 = -\frac{x^{-2}}{2} + C_2\).

Combine the results:

\(\int \left( x^3 + \frac{1}{x^3} \right) \, dx = \frac{x^4}{4} - \frac{x^{-2}}{2} + c\), where \(c = C_1 + C_2\).

First, expand the integrand:

\(\left( x + \frac{1}{x} \right)^2 = x^2 + 2 \cdot x \cdot \frac{1}{x} + \left( \frac{1}{x} \right)^2 = x^2 + 2 + \frac{1}{x^2}\).

Now, integrate each term separately:

\(\int x^2 \, dx = \frac{x^3}{3} + C_1\)

\(\int 2 \, dx = 2x + C_2\)

\(\int \frac{1}{x^2} \, dx = \int x^{-2} \, dx = -x^{-1} + C_3 = -\frac{1}{x} + C_3\)

Combine the results:

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

Let \(u = 3x + 1\). Then \(du = 3 \, dx\) or \(dx = \frac{du}{3}\).

The limits of integration change from \(x = 0\) to \(u = 1\) and from \(x = 1\) to \(u = 4\).

Thus, the integral becomes:

\(\int_{1}^{4} \sqrt{u} \cdot \frac{1}{3} \, du = \frac{1}{3} \int_{1}^{4} u^{0.5} \, du\).

Integrate \(u^{0.5}\):

\(\frac{1}{3} \left[ \frac{u^{1.5}}{1.5} \right]_{1}^{4} = \frac{1}{3} \cdot \frac{2}{3} \left[ u^{1.5} \right]_{1}^{4}\).

Evaluate at the bounds:

\(\frac{2}{9} \left[ 4^{1.5} - 1^{1.5} \right] = \frac{2}{9} \left[ 8 - 1 \right] = \frac{2}{9} \cdot 7 = \frac{14}{9}\).

Thus, the answer is \(\frac{14}{9}\) or 1.56.