What is the integral of (4x^2-2x-2) ?

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

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integral of (4x^2-2x-2)

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

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

Solution steps

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

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

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

\int 2 d x = 2 x

= \frac{4 x ^{3}}{3} - x^{2} - 2 x

Add a constant to the solution

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

\int 25 + x d x

\frac{d}{d x} (e^{3 x} + e^{- 2 x})

y^{^{''}} + 6 y^{^{'}} + 15 y = 0

\int (3 t^{4} - 8) t^{3} d t

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

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