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

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

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integral of (3x^2+4x+2)

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

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

Solution steps

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

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

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

\int 2 d x = 2 x

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{48}{( 1 - 2 x ) ^{4}})

\frac{d}{d x} (ln (x + 2 + x^{2}))

x y^{2} y^{^{'}} = x^{3} + y^{3}

\int_{0}^{\frac{π}{6}} ln (sec (x)) d x

y = x, y = x^{2}, x = 2

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