What is the integral of-2x^3(-5x^2-x+12) ?

The integral of-2x^3(-5x^2-x+12) is -2(-(5x^6)/6-(x^5)/5+3x^4)+C

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integral of-2x^3(-5x^2-x+12)

\int - 2 x^{3} (- 5 x^{2} - x + 12) d x

- 2 (- \frac{5 x ^{6}}{6} - \frac{x ^{5}}{5} + 3 x^{4}) + C

Solution steps

\int - 2 x^{3} (- 5 x^{2} - x + 12) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 2 \cdot \int x^{3} (- 5 x^{2} - x + 12) d x

Expand

x^{3} (- 5 x^{2} - x + 12) : - 5 x^{5} - x^{4} + 12 x^{3}

= - 2 \cdot \int - 5 x^{5} - x^{4} + 12 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

= - 2 (- \int 5 x^{5} d x - \int x^{4} d x + \int 12 x^{3} d x)

\int 5 x^{5} d x = \frac{5 x ^{6}}{6}

\int x^{4} d x = \frac{x ^{5}}{5}

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

= - 2 (- \frac{5 x ^{6}}{6} - \frac{x ^{5}}{5} + 3 x^{4})

Add a constant to the solution

= - 2 (- \frac{5 x ^{6}}{6} - \frac{x ^{5}}{5} + 3 x^{4}) + C

\frac{\partial}{\partial z} (- 2 e^{x} cos (yz))

\int x^{3} (1 + x^{4})^{3} d x

\frac{d}{d x} (sin (x + 2))

\int_{- \infty}^{0} \frac{1}{4 - 3 x} d x

\frac{s}{( s - 3 ) ^{3}}

What is the integral of-2x^3(-5x^2-x+12) ?
The integral of-2x^3(-5x^2-x+12) is -2(-(5x^6)/6-(x^5)/5+3x^4)+C