What is the integral of-19x^2-2x+8 ?

The integral of-19x^2-2x+8 is -(19x^3)/3-x^2+8x+C

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integral of-19x^2-2x+8

\int - 19 x^{2} - 2 x + 8 d x

- \frac{19 x ^{3}}{3} - x^{2} + 8 x + C

Solution steps

\int - 19 x^{2} - 2 x + 8 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 19 x^{2} d x - \int 2 x d x + \int 8 d x

\int 19 x^{2} d x = \frac{19 x ^{3}}{3}

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

\int 8 d x = 8 x

= - \frac{19 x ^{3}}{3} - x^{2} + 8 x

Add a constant to the solution

= - \frac{19 x ^{3}}{3} - x^{2} + 8 x + C

\frac{\partial}{\partial x} (1 + x^{3} + y^{3})

g (x) = - \frac{9}{( 6 x + 8 ) ^{\frac{3}{2}}}

\int 5^{3 x} d x

\frac{d}{d x} (cos^{2} (x) d x)

What is the integral of-19x^2-2x+8 ?
The integral of-19x^2-2x+8 is -(19x^3)/3-x^2+8x+C