What is the integral of (6x^2+8x+3) ?

The integral of (6x^2+8x+3) is 2x^3+4x^2+3x+C

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integral of (6x^2+8x+3)

\int (6 x^{2} + 8 x + 3) d x

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

Solution steps

\int 6 x^{2} + 8 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 6 x^{2} d x + \int 8 x d x + \int 3 d x

\int 6 x^{2} d x = 2 x^{3}

\int 8 x d x = 4 x^{2}

\int 3 d x = 3 x

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

Add a constant to the solution

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

y^{^{'}} = 8 x^{3} e^{- 2 y}

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

\frac{( 2 + s )}{s ^{2} + 2 s + 2}

\int \frac{2}{sin ^{2} ( x )} d x

5 x^{4} - x - \frac{2}{x}

What is the integral of (6x^2+8x+3) ?
The integral of (6x^2+8x+3) is 2x^3+4x^2+3x+C