What is the integral of e^{2x}+e^{-2x} ?

The integral of e^{2x}+e^{-2x} is 1/2 e^{2x}-1/2 e^{-2x}+C

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integral of e^{2x}+e^{-2x}

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

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

Solution steps

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

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

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

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

Add a constant to the solution

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

h \to 0 lim (2.5 h^{2} + 6 h)

3 t + 2 e^{3 t}

\frac{\partial}{\partial x} ((sin (y)) u (x, π))

\int_{0}^{\frac{3}{2}} \frac{1}{36 + 16 x ^{2}} d x

\frac{1}{( 16 x ^{2} )}, x, \frac{x}{6}

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