What is the integral of (e^{1-3x}+2cos(x/9)) ?

The integral of (e^{1-3x}+2cos(x/9)) is -1/3 e^{1-3x}+18sin(x/9)+C

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integral of (e^{1-3x}+2cos(x/9))

\int (e^{1 - 3 x} + 2 cos (\frac{x}{9})) d x

- \frac{1}{3} e^{1 - 3 x} + 18 sin (\frac{x}{9}) + C

Solution steps

\int e^{1 - 3 x} + 2 cos (\frac{x}{9}) 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 e^{1 - 3 x} d x + \int 2 cos (\frac{x}{9}) d x

\int e^{1 - 3 x} d x = - \frac{1}{3} e^{1 - 3 x}

\int 2 cos (\frac{x}{9}) d x = 18 sin (\frac{x}{9})

= - \frac{1}{3} e^{1 - 3 x} + 18 sin (\frac{x}{9})

Add a constant to the solution

= - \frac{1}{3} e^{1 - 3 x} + 18 sin (\frac{x}{9}) + C

What is the integral of (e^{1-3x}+2cos(x/9)) ?
The integral of (e^{1-3x}+2cos(x/9)) is -1/3 e^{1-3x}+18sin(x/9)+C