What is the integral of (-6x^2+4x-4xcos(x)) ?

The integral of (-6x^2+4x-4xcos(x)) is -2x^3+2x^2-4(xsin(x)+cos(x))+C

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integral of (-6x^2+4x-4xcos(x))

\int (- 6 x^{2} + 4 x - 4 x cos (x)) d x

- 2 x^{3} + 2 x^{2} - 4 (x sin (x) + cos (x)) + C

Solution steps

\int - 6 x^{2} + 4 x - 4 x cos (x) 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 4 x d x - \int 4 x cos (x) d x

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

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

\int 4 x cos (x) d x = 4 (x sin (x) + cos (x))

= - 2 x^{3} + 2 x^{2} - 4 (x sin (x) + cos (x))

Add a constant to the solution

= - 2 x^{3} + 2 x^{2} - 4 (x sin (x) + cos (x)) + C

\int_{0}^{8} (8 - x) d x

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

\int \frac{e ^{2 x}}{2 + e ^{x}} d x

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

\frac{\partial}{\partial x} (x^{2} - 8 x + y^{2} \cdot ln (x - 1))

What is the integral of (-6x^2+4x-4xcos(x)) ?
The integral of (-6x^2+4x-4xcos(x)) is -2x^3+2x^2-4(xsin(x)+cos(x))+C