What is the integral of 4x^3-3x^2+4x-3 ?

The integral of 4x^3-3x^2+4x-3 is x^4-x^3+2x^2-3x+C

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integral of 4x^3-3x^2+4x-3

\int 4 x^{3} - 3 x^{2} + 4 x - 3 d x

x^{4} - x^{3} + 2 x^{2} - 3 x + C

Solution steps

\int 4 x^{3} - 3 x^{2} + 4 x - 3 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 4 x^{3} d x - \int 3 x^{2} d x + \int 4 x d x - \int 3 d x

\int 4 x^{3} d x = x^{4}

\int 3 x^{2} d x = x^{3}

\int 4 x d x = 2 x^{2}

\int 3 d x = 3 x

= x^{4} - x^{3} + 2 x^{2} - 3 x

Add a constant to the solution

= x^{4} - x^{3} + 2 x^{2} - 3 x + C

What is the integral of 4x^3-3x^2+4x-3 ?
The integral of 4x^3-3x^2+4x-3 is x^4-x^3+2x^2-3x+C