If `x=(6y+4)(3y^(2)+4y+3)` then `intxdy` will be `:`

Options

A. `(1)/(3x^(2)+4y+3)`
B. `((3y^(2)+4y+3)^(2))/(2)+C`
C. `(3y^(2)+4y+3)`
D. `((6y+4))/((3y^(2)+4y+3))`

Explanation

To solve the problem, we need to find the integral of the expression given by \( x = (6y + 4)(3y^2 + 4y + 3) \) with respect to \( y \). Let's break this down step by step.

### Step 1: Set up the integral
We start with the expression for \( x \):
\[ x = (6y + 4)(3y^2 + 4y + 3) \]
We need to compute the integral:
\[ I = \int x \, dy = \int (6y + 4)(3y^2 + 4y + 3) \, dy \]

### Step 2: Use substitution
To simplify the integration, we can use substitution. Let's set:
\[ t = 3y^2 + 4y + 3 \]
Now, we need to find \( dt \) in terms of \( dy \). We differentiate \( t \):
\[ \frac{dt}{dy} = 6y + 4 \]
Thus, we can express \( dt \) as:
\[ dt = (6y + 4) \, dy \]
This means:
\[ (6y + 4) \, dy = dt \]

### Step 3: Rewrite the integral
Now, we can rewrite the integral \( I \) in terms of \( t \):
\[ I = \int t \, dt \]

### Step 4: Integrate
The integral of \( t \) with respect to \( t \) is:
\[ I = \frac{t^2}{2} + C \]
where \( C \) is the constant of integration.

### Step 5: Substitute back for \( t \)
Now we substitute back \( t = 3y^2 + 4y + 3 \):
\[ I = \frac{(3y^2 + 4y + 3)^2}{2} + C \]

### Final Result
Thus, the integral \( \int x \, dy \) is:
\[ \int x \, dy = \frac{(3y^2 + 4y + 3)^2}{2} + C \]