What is the integral of (e^{2m}+e^{-2m}) ?

The integral of (e^{2m}+e^{-2m}) is 1/2 e^{2m}-1/2 e^{-2m}+C

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integral of (e^{2m}+e^{-2m})

\int (e^{2 m} + e^{- 2 m}) d m

\frac{1}{2} e^{2 m} - \frac{1}{2} e^{- 2 m} + C

Solution steps

\int e^{2 m} + e^{- 2 m} d m

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^{2 m} d m + \int e^{- 2 m} d m

\int e^{2 m} d m = \frac{1}{2} e^{2 m}

\int e^{- 2 m} d m = - \frac{1}{2} e^{- 2 m}

= \frac{1}{2} e^{2 m} - \frac{1}{2} e^{- 2 m}

Add a constant to the solution

= \frac{1}{2} e^{2 m} - \frac{1}{2} e^{- 2 m} + C

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What is the integral of (e^{2m}+e^{-2m}) ?
The integral of (e^{2m}+e^{-2m}) is 1/2 e^{2m}-1/2 e^{-2m}+C