What is the integral of 3e^x-9cos(x)+C ?

The integral of 3e^x-9cos(x)+C is Cx+3e^x-9sin(x)+C

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integral of 3e^x-9cos(x)+C

\int 3 e^{x} - 9 cos (x) + C d x

C x + 3 e^{x} - 9 sin (x) + C

Solution steps

\int 3 e^{x} - 9 cos (x) + C 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 C d x + \int 3 e^{x} d x - \int 9 cos (x) d x

\int C d x = C x

\int 3 e^{x} d x = 3 e^{x}

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

= C x + 3 e^{x} - 9 sin (x)

Add a constant to the solution

= C x + 3 e^{x} - 9 sin (x) + C

4 x^{3}

\frac{d y}{d x} = tan (x) + sec (x)

\frac{cos ( x )}{2 sin ( x )}

\frac{\partial}{\partial v} (4 u sin (v))

\int x x - 15 d x

What is the integral of 3e^x-9cos(x)+C ?
The integral of 3e^x-9cos(x)+C is Cx+3e^x-9sin(x)+C