What is the integral of (4x^3-4x+6) ?

The integral of (4x^3-4x+6) is x^4-2x^2+6x+C

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integral of (4x^3-4x+6)

\int (4 x^{3} - 4 x + 6) d x

x^{4} - 2 x^{2} + 6 x + C

Solution steps

\int 4 x^{3} - 4 x + 6 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 4 x d x + \int 6 d x

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

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

\int 6 d x = 6 x

= x^{4} - 2 x^{2} + 6 x

Add a constant to the solution

= x^{4} - 2 x^{2} + 6 x + C

\int x^{2} - 6 x + 9 d x

\int - x \cdot cos (x) d x

\int \frac{8 x - 1}{( x - 2 ) ( x + 3 )} d x

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

\int \frac{1}{sin ( x ) 1 + cos ( x )} d x

What is the integral of (4x^3-4x+6) ?
The integral of (4x^3-4x+6) is x^4-2x^2+6x+C