The indefinite integral of $$2x^3-x^2+3$$ is:

To solve this indefinite integral, we must apply the addition rule of integrals which states that the integral of a sum is the sum of the individual integrals. Therefore, we can write the expression of the indefinite integral as:

$\int (2x^3-x^2+3)dx = \int \left(2x^3\right) -\int \left(x^2\right)+\int \left(3\right)$

When integrating expressions with exponents, we simply add +1 to the exponent and then divide the expression by the new exponent. We can think of the last term as $3 = 3x^0$. Solving each individual integral, we get:

$\int (2x^3-x^2+3)dx =\frac{ 2x^4}{4} -\frac{x^3}{3}+3x = \frac{x^4}{2}-\frac{x^3}{3}+3x$

As with solving all indefinite integrals, we must add a constant C term to the expression to arrive at the general solution:

$\int (2x^3-x^2+3)dx = \frac{x^4}{2}-\frac{x^3}{3}+3x+C$