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

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

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

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

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

Solution steps

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

Expand

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

\int \frac{e ^{2 x}}{e ^{2 x} + e ^{- 2 x}} d x = \frac{1}{4} ln e^{4 x} + 1

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

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

Add a constant to the solution

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

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