To solve the integral \( I = \int (\sin(x^2) + 2x^2 \cos(x^2)) \, dx \) and express it in the form \( I = x h(x) + C \), we will follow these steps:
### Step 1: Rewrite the Integral
We start with the integral:
\[
I = \int (\sin(x^2) + 2x^2 \cos(x^2)) \, dx
\]
We can separate the integral into two parts:
\[
I = \int \sin(x^2) \, dx + \int 2x^2 \cos(x^2) \, dx
\]
### Step 2: Identify the Derivative Form
Notice that the second term \( 2x^2 \cos(x^2) \) can be recognized as a product of a function and its derivative. Specifically, if we let \( u = x^2 \), then \( du = 2x \, dx \). We can rewrite the second integral:
\[
\int 2x^2 \cos(x^2) \, dx = \int x \cdot 2x \cos(x^2) \, dx = \int x \cdot \cos(u) \, du
\]
### Step 3: Apply Integration by Parts
Using integration by parts on the second integral, we let:
- \( f = x \) and \( dg = \cos(x^2) \, dx \)
- Then \( df = dx \) and \( g = \frac{1}{2} \sin(x^2) \)
Thus, we have:
\[
\int x \cos(x^2) \, dx = x \cdot \frac{1}{2} \sin(x^2) - \int \frac{1}{2} \sin(x^2) \, dx
\]
### Step 4: Combine the Results
Now, substituting back into our integral \( I \):
\[
I = \int \sin(x^2) \, dx + x \cdot \sin(x^2) - \frac{1}{2} \int \sin(x^2) \, dx
\]
Combining the terms gives:
\[
I = \frac{1}{2} \int \sin(x^2) \, dx + x \cdot \sin(x^2) + C
\]
### Step 5: Express in the Required Form
From the expression above, we can identify:
\[
I = x h(x) + C
\]
where \( h(x) = \frac{1}{2} \sin(x^2) + \sin(x^2) = \frac{3}{2} \sin(x^2) \).
### Step 6: Determine the Range of \( h(x) \)
The range of \( h(x) = \frac{3}{2} \sin(x^2) \) is determined by the range of \( \sin(x^2) \), which is \([-1, 1]\). Thus:
\[
h(x) \text{ ranges from } \frac{3}{2} \cdot (-1) = -\frac{3}{2} \text{ to } \frac{3}{2} \cdot 1 = \frac{3}{2}.
\]
So, we have \( a = -\frac{3}{2} \) and \( b = \frac{3}{2} \).
### Step 7: Calculate \( a + 2b \)
Now, we can compute:
\[
a + 2b = -\frac{3}{2} + 2 \cdot \frac{3}{2} = -\frac{3}{2} + 3 = \frac{3}{2}.
\]
Thus, the final answer is:
\[
\boxed{\frac{3}{2}}.
\]
