What is the integral of (e^{4x}+2)/(e^{4x)+e^{-4x}} ?

The integral of (e^{4x}+2)/(e^{4x)+e^{-4x}} is 1/8 ln|e^{8x}+1|-1/2 arctan(e^{-4x})+C

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integral of (e^{4x}+2)/(e^{4x)+e^{-4x}}

\int \frac{e ^{4 x} + 2}{e ^{4 x} + e ^{- 4 x}} d x

\frac{1}{8} ln e^{8 x} + 1 - \frac{1}{2} arctan (e^{- 4 x}) + C

Solution steps

\int \frac{e ^{4 x} + 2}{e ^{4 x} + e ^{- 4 x}} d x

Expand

\frac{e ^{4 x} + 2}{e ^{4 x} + e ^{- 4 x}} : \frac{e ^{4 x}}{e ^{4 x} + e ^{- 4 x}} + \frac{2}{e ^{4 x} + e ^{- 4 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 \frac{e ^{4 x}}{e ^{4 x} + e ^{- 4 x}} d x + \int \frac{2}{e ^{4 x} + e ^{- 4 x}} d x

\int \frac{e ^{4 x}}{e ^{4 x} + e ^{- 4 x}} d x = \frac{1}{8} ln e^{8 x} + 1

\int \frac{2}{e ^{4 x} + e ^{- 4 x}} d x = - \frac{1}{2} arctan (e^{- 4 x})

= \frac{1}{8} ln e^{8 x} + 1 - \frac{1}{2} arctan (e^{- 4 x})

Add a constant to the solution

= \frac{1}{8} ln e^{8 x} + 1 - \frac{1}{2} arctan (e^{- 4 x}) + C

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What is the integral of (e^{4x}+2)/(e^{4x)+e^{-4x}} ?
The integral of (e^{4x}+2)/(e^{4x)+e^{-4x}} is 1/8 ln|e^{8x}+1|-1/2 arctan(e^{-4x})+C