Question

find the integral
\int \left(3x^{2} - 4x + 4\
ight) dx

\int \left(3x^{2} - 4x + 4\
ight) dx = \square

Explanation:

Step1: Integrate term by term

Apply power rule: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$
$\int 3x^2 dx = 3 \cdot \frac{x^3}{3} = x^3$
$\int -4x dx = -4 \cdot \frac{x^2}{2} = -2x^2$
$\int 4 dx = 4x$

Step2: Combine terms + constant

Sum results and add $C$
$x^3 - 2x^2 + 4x + C$

Answer:

$x^3 - 2x^2 + 4x + C$