What is the integral of (e^{6x}+4)/(e^{6x)+e^{-6x}} ?

The integral of (e^{6x}+4)/(e^{6x)+e^{-6x}} is 1/12 ln|e^{12x}+1|-2/3 arctan(e^{-6x})+C

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integral of (e^{6x}+4)/(e^{6x)+e^{-6x}}

\int \frac{e ^{6 x} + 4}{e ^{6 x} + e ^{- 6 x}} d x

\frac{1}{12} ln e^{12 x} + 1 - \frac{2}{3} arctan (e^{- 6 x}) + C

Solution steps

\int \frac{e ^{6 x} + 4}{e ^{6 x} + e ^{- 6 x}} d x

Expand

\frac{e ^{6 x} + 4}{e ^{6 x} + e ^{- 6 x}} : \frac{e ^{6 x}}{e ^{6 x} + e ^{- 6 x}} + \frac{4}{e ^{6 x} + e ^{- 6 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 ^{6 x}}{e ^{6 x} + e ^{- 6 x}} d x + \int \frac{4}{e ^{6 x} + e ^{- 6 x}} d x

\int \frac{e ^{6 x}}{e ^{6 x} + e ^{- 6 x}} d x = \frac{1}{12} ln e^{12 x} + 1

\int \frac{4}{e ^{6 x} + e ^{- 6 x}} d x = - \frac{2}{3} arctan (e^{- 6 x})

= \frac{1}{12} ln e^{12 x} + 1 - \frac{2}{3} arctan (e^{- 6 x})

Add a constant to the solution

= \frac{1}{12} ln e^{12 x} + 1 - \frac{2}{3} arctan (e^{- 6 x}) + C

What is the integral of (e^{6x}+4)/(e^{6x)+e^{-6x}} ?
The integral of (e^{6x}+4)/(e^{6x)+e^{-6x}} is 1/12 ln|e^{12x}+1|-2/3 arctan(e^{-6x})+C