To solve the integral \( \int (x^6 + x^4 + x^2)(2x^4 + 3x^2 + 6)^{1/2} \, dx \), we will follow a systematic approach.
### Step 1: Rewrite the Integral
We start with the integral:
\[
I = \int (x^6 + x^4 + x^2)(2x^4 + 3x^2 + 6)^{1/2} \, dx
\]
### Step 2: Factor out \( x^2 \)
Notice that we can factor \( x^2 \) from the polynomial:
\[
x^6 + x^4 + x^2 = x^2(x^4 + x^2 + 1)
\]
Thus, we can rewrite the integral as:
\[
I = \int x^2 (x^4 + x^2 + 1)(2x^4 + 3x^2 + 6)^{1/2} \, dx
\]
### Step 3: Substitution
Let us make the substitution:
\[
t = 2x^6 + 3x^4 + 6x^2
\]
Now, we need to differentiate \( t \) with respect to \( x \):
\[
\frac{dt}{dx} = 12x^5 + 12x^3 + 12x = 12x(x^4 + x^2 + 1)
\]
Thus, we have:
\[
dt = 12x(x^4 + x^2 + 1) \, dx \quad \Rightarrow \quad dx = \frac{dt}{12x(x^4 + x^2 + 1)}
\]
### Step 4: Substitute in the Integral
Now substituting \( t \) and \( dx \) into the integral:
\[
I = \int x^2 (x^4 + x^2 + 1)(2x^4 + 3x^2 + 6)^{1/2} \cdot \frac{dt}{12x(x^4 + x^2 + 1)}
\]
The \( (x^4 + x^2 + 1) \) terms cancel out:
\[
I = \int \frac{x^2}{12x} (2x^4 + 3x^2 + 6)^{1/2} \, dt = \frac{1}{12} \int (2x^4 + 3x^2 + 6)^{1/2} \, dt
\]
### Step 5: Change of Variables
Now, we can express \( (2x^4 + 3x^2 + 6)^{1/2} \) in terms of \( t \):
\[
I = \frac{1}{12} \int t^{1/2} \, dt
\]
### Step 6: Evaluate the Integral
The integral of \( t^{1/2} \) is:
\[
\int t^{1/2} \, dt = \frac{t^{3/2}}{3/2} = \frac{2}{3} t^{3/2}
\]
Thus, we have:
\[
I = \frac{1}{12} \cdot \frac{2}{3} t^{3/2} + C_1 = \frac{1}{18} t^{3/2} + C_1
\]
### Step 7: Substitute Back for \( t \)
Substituting back for \( t \):
\[
t = 2x^6 + 3x^4 + 6x^2
\]
So, we get:
\[
I = \frac{1}{18} (2x^6 + 3x^4 + 6x^2)^{3/2} + C_1
\]
### Step 8: Identify Constants
From the expression, we can identify:
- \( k = \frac{1}{18} \)
- \( A = 2 \)
- \( B = 3 \)
- \( C = 6 \)
- \( p = \frac{3}{2} \)
### Final Answer
Thus, we have:
\[
I = k(Ax^6 + Bx^4 + Cx^2)^p + C_1
\]
where \( k = \frac{1}{18}, A = 2, B = 3, C = 6, p = \frac{3}{2} \).
