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

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

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integral of (6x^2+8x-sin(x))

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

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

Solution steps

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

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

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

\int sin (x) d x = - cos (x)

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

Simplify

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

Add a constant to the solution

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

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