To solve the given integral $\int 6 x^3 e^{x^2 + 3} \, dx$ using integration by parts, we will first identify the functions $u$ and $dv$.

Given: $\int 6 x^3 e^{x^2 + 3} \, dx$

We let $u = 6 x^2$ and $dv = x e^{x^2 + 3} \, dx$.

Next, we need to find $du$ and $v$:

$du = \frac{d}{dx}(6 x^2) \, dx = 12 x \, dx$ $dv = x e^{x^2 + 3} \, dx$

To find $v$, we integrate $dv$:

$v = \int x e^{x^2 + 3} \, dx$

Let $w = x^2 + 3$, so $dw = 2x \, dx$ and $\frac{dw}{2} = x \, dx$.

Thus,

$v = \int x e^{x^2 + 3} \, dx = \int e^w \frac{dw}{2} = \frac{1}{2} \int e^w \, dw = \frac{1}{2} e^w = \frac{1}{2} e^{x^2 + 3}$

So, now we have:

$u = 6 x^2$ $dv = x e^{x^2 + 3} \, dx$ $du = 12 x \, dx$ $v = \frac{1}{2} e^{x^2 + 3}$

Using integration by parts formula $\int u \, dv = uv - \int v \, du$, we get:

$\int 6 x^3 e^{x^2 + 3} \, dx = 6 x^2 \cdot \frac{1}{2} e^{x^2 + 3} - \int \left( \frac{1}{2} e^{x^2 + 3} \right) \cdot (12 x) \, dx$

Simplifying:

$= 3 x^2 e^{x^2 + 3} - 6 \int x e^{x^2 + 3} \, dx$

We already know $\int x e^{x^2 + 3} \, dx = \frac{1}{2} e^{x^2 + 3}$, so:

$= 3 x^2 e^{x^2 + 3} - 6 \cdot \frac{1}{2} e^{x^2 + 3}$ $= 3 x^2 e^{x^2 + 3} - 3 e^{x^2 + 3}$ $= e^{x^2 + 3} (3 x^2 - 3)$

The integral simplifies to:

$\int 6 x^3 e^{x^2 + 3} \, dx = e^{x^2 + 3} (3 x^2 - 3) + C$

Now, we need to compare this with the options given and identify $h(x)$. Here, $h(x)$ represents the term we found within the integral $e^{x^2 + 3}$.

Comparing with the options, we find:

a.) $6 e^{x^2 + 3}$ b.) $3 e^{-3} e^{x^2}$ c.) $\frac{3 e^{x^2 + 6}}{e^{3}}$ d.) $2 e^{x^2 + 3}$

The correct answer is:

$\boxed{6 e^{x^2 + 3}}$

Thus, $h(x) = 6 e^{x^2 + 3}$.