What is the integral of 8x^3+9sin(x) ?

The integral of 8x^3+9sin(x) is 2x^4-9cos(x)+C

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integral of 8x^3+9sin(x)

\int 8 x^{3} + 9 sin (x) d x

2 x^{4} - 9 cos (x) + C

Solution steps

\int 8 x^{3} + 9 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 8 x^{3} d x + \int 9 sin (x) d x

\int 8 x^{3} d x = 2 x^{4}

\int 9 sin (x) d x = - 9 cos (x)

= 2 x^{4} - 9 cos (x)

Add a constant to the solution

= 2 x^{4} - 9 cos (x) + C

- \frac{12}{x ^{4}}

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (t)

x \to 3 lim (3 x^{4})

arctan (x), 1

\frac{\partial}{\partial z} (- 5 e^{3 x + 4 y} sin (5 z))

What is the integral of 8x^3+9sin(x) ?
The integral of 8x^3+9sin(x) is 2x^4-9cos(x)+C