To evaluate the line integral of the vector field $\mathbf{F}(x, y, z) = 3x^2z \, \mathbf{i} + z^2 \, \mathbf{j} + (x^3 + 2yz) \, \mathbf{k}$ along the curve $C$ parameterized by $\mathbf{r}(t) = \ln(t) \, \mathbf{i} + \frac{t^{3/2}}{\ln(2)} \, \mathbf{j} + t \cos(\pi t) \, \mathbf{k}$ for $1 \leq t \leq 4$, we proceed as follows:

Step 1: Compute $\mathbf{F}(\mathbf{r}(t))$

First, we substitute $x = \ln(t)$, $y = \frac{t^{3/2}}{\ln(2)}$, and $z = t \cos(\pi t)$ into $\mathbf{F}(x, y, z)$:

1. The $i$-component of $\mathbf{F}$ is $3x^2z = 3(\ln(t))^2 \cdot t \cos(\pi t)$.
2. The $j$-component of $\mathbf{F}$ is $z^2 = (t \cos(\pi t))^2 = t^2 \cos^2(\pi t)$.
3. The $k$-component of $\mathbf{F}$ is $x^3 + 2yz = (\ln(t))^3 + 2 \cdot \frac{t^{3/2}}{\ln(2)} \cdot t \cos(\pi t)$.

So, $\mathbf{F}(\mathbf{r}(t)) = 3(\ln(t))^2 \, t \cos(\pi t) \, \mathbf{i} + t^2 \cos^2(\pi t) \, \mathbf{j} + \left((\ln(t))^3 + 2 \frac{t^{5/2} \cos(\pi t)}{\ln(2)}\right) \, \mathbf{k}.$

Step 2: Compute $\frac{d\mathbf{r}}{dt}$

Differentiate each component of $\mathbf{r}(t)$ with respect to $t$:

1. For the $i$-component, $\frac{d}{dt} \ln(t) = \frac{1}{t}$.
2. For the $j$-component, $\frac{d}{dt} \frac{t^{3/2}}{\ln(2)} = \frac{3}{2} \frac{t^{1/2}}{\ln(2)}$.
3. For the $k$-component, $\frac{d}{dt} (t \cos(\pi t)) = \cos(\pi t) - \pi t \sin(\pi t)$.

Thus, $\frac{d\mathbf{r}}{dt} = \frac{1}{t} \, \mathbf{i} + \frac{3}{2} \frac{t^{1/2}}{\ln(2)} \, \mathbf{j} + \left(\cos(\pi t) - \pi t \sin(\pi t)\right) \, \mathbf{k}.$

Step 3: Compute $\mathbf{F}(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt}$

The dot product $\mathbf{F}(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt}$ is given by: $\mathbf{F}(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt} = \left(3(\ln(t))^2 \, t \cos(\pi t)\right) \cdot \frac{1}{t} + \left(t^2 \cos^2(\pi t)\right) \cdot \frac{3}{2} \frac{t^{1/2}}{\ln(2)} + \left((\ln(t))^3 + 2 \frac{t^{5/2} \cos(\pi t)}{\ln(2)}\right) \cdot \left(\cos(\pi t) - \pi t \sin(\pi t)\right).$

Simplify each term individually:

1. The first term: $3(\ln(t))^2 \cos(\pi t).$
2. The second term: $\frac{3}{2} \frac{t^{5/2} \cos^2(\pi t)}{\ln(2)}.$
3. The third term: $\left((\ln(t))^3 + 2 \frac{t^{5/2} \cos(\pi t)}{\ln(2)}\right) \left(\cos(\pi t) - \pi t \sin(\pi t)\right).$

Step 4: Integrate from $t = 1$ to $t = 4$

Now we need to integrate the resulting expression with respect to $t$ over the interval $[1, 4]$:

$\int_{1}^{4} \left(3(\ln(t))^2 \cos(\pi t) + \frac{3}{2} \frac{t^{5/2} \cos^2(\pi t)}{\ln(2)} + \left((\ln(t))^3 + 2 \frac{t^{5/2} \cos(\pi t)}{\ln(2)}\right) \left(\cos(\pi t) - \pi t \sin(\pi t)\right)\right) \, dt.$

This integral is quite complex and would generally be evaluated using numerical integration techniques.

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Tip: When solving line integrals involving complex functions, substituting and simplifying before integrating can sometimes reveal easier ways to handle the calculation.