What is the integral of x^2-4cos(x)+8 ?

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

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integral of x^2-4cos(x)+8

\int x^{2} - 4 cos (x) + 8 d x

\frac{x ^{3}}{3} - 4 sin (x) + 8 x + C

Solution steps

\int x^{2} - 4 cos (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 x^{2} d x - \int 4 cos (x) d x + \int 8 d x

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

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

\int 8 d x = 8 x

= \frac{x ^{3}}{3} - 4 sin (x) + 8 x

Add a constant to the solution

= \frac{x ^{3}}{3} - 4 sin (x) + 8 x + C

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\frac{\partial}{\partial x} (100 e^{(2 x + 3 y)})

\int e^{x} 6 + e^{x} d x

(\frac{1}{2} 3 x)^{^{'}}

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