What is the integral of (12x^3-4x-2) ?

The integral of (12x^3-4x-2) is 3x^4-2x^2-2x+C

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integral of (12x^3-4x-2)

\int (12 x^{3} - 4 x - 2) d x

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

Solution steps

\int 12 x^{3} - 4 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 12 x^{3} d x - \int 4 x d x - \int 2 d x

\int 12 x^{3} d x = 3 x^{4}

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

\int 2 d x = 2 x

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

Add a constant to the solution

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

\int \frac{1}{\frac{1}{v}} d v

\frac{e ^{- 2 s}}{s ^{2} + 1}

5 x^{2} + y^{2} = 9

\int x^{3} x + 1 d x

y^{^{''}} - 2 y^{^{'}} - 24 y = 0

What is the integral of (12x^3-4x-2) ?
The integral of (12x^3-4x-2) is 3x^4-2x^2-2x+C