What is the integral of 200e^{0.1x}+120e^{-0.3x} ?

The integral of 200e^{0.1x}+120e^{-0.3x} is 2000e^{0.1x}-399.99999…e^{-0.3x}+C

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integral of 200e^{0.1x}+120e^{-0.3x}

\int 200 e^{0.1 x} + 120 e^{- 0.3 x} d x

2000 e^{0.1 x} - 399.99999 \dots e^{- 0.3 x} + C

Solution steps

\int 200 e^{0.1 x} + 120 e^{- 0.3 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 200 e^{0.1 x} d x + \int 120 e^{- 0.3 x} d x

\int 200 e^{0.1 x} d x = 2000 e^{0.1 x}

\int 120 e^{- 0.3 x} d x = - 399.99999 \dots e^{- 0.3 x}

= 2000 e^{0.1 x} - 399.99999 \dots e^{- 0.3 x}

Add a constant to the solution

= 2000 e^{0.1 x} - 399.99999 \dots e^{- 0.3 x} + C

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What is the integral of 200e^{0.1x}+120e^{-0.3x} ?
The integral of 200e^{0.1x}+120e^{-0.3x} is 2000e^{0.1x}-399.99999…e^{-0.3x}+C