What is the integral of (x^2 + 3x + 2) dx?

What is the integral of (x^2 + 3x + 2) dx?

• A. (1/3)x^3 + (3/2)x^2 + 2x + C
• B. (1/4)x^4 + (3/3)x^2 + 2x + C
• C. (2/3)x^3 + (3/2)x^2 + C
• D. (1/2)x^3 + (3/4)x^2 + 2x + C

Answer: (A)

To find the integral of the polynomial \( x^2 + 3x + 2 \), we apply the standard rules of integration for polynomials. Each term of the polynomial can be integrated separately. Starting with the first term \( x^2 \), we use the power rule, which states that the integral of \( x^n \) is \( \frac{1}{n+1}x^{n+1} \) for \( n \neq -1 \). Here, \( n = 2 \): \[ \int x^2 \, dx = \frac{1}{2 + 1}x^{2 + 1} = \frac{1}{3}x^3. \] Next, we integrate the second term \( 3x \): \[ \int 3x \, dx = 3 \cdot \frac{1}{1 + 1}x^{1 + 1} = 3 \cdot \frac{1}{2}x^2 = \frac{3}{2}x^2. \] Lastly, we integrate the constant term \( 2 \): \[ \int 2 \, dx = 2x

To find the integral of the polynomial ( x^2 + 3x + 2 ), we apply the standard rules of integration for polynomials. Each term of the polynomial can be integrated separately.

Starting with the first term ( x^2 ), we use the power rule, which states that the integral of ( x^n ) is ( \frac{1}{n+1}x^{n+1} ) for ( n \neq -1 ). Here, ( n = 2 ):

[

\int x^2 , dx = \frac{1}{2 + 1}x^{2 + 1} = \frac{1}{3}x^3.

]

Next, we integrate the second term ( 3x ):

[

\int 3x , dx = 3 \cdot \frac{1}{1 + 1}x^{1 + 1} = 3 \cdot \frac{1}{2}x^2 = \frac{3}{2}x^2.

]

Lastly, we integrate the constant term ( 2 ):

[

\int 2 , dx = 2x