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

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

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integral of 1/(e^x+2e^{-x)+2}

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

- arctan (2 e^{- x} + 1) + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{1 + 2 u ^{2} + 2 u} d u

= \int - \frac{1}{1 + 2 u ^{2} + 2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{1 + 2 u ^{2} + 2 u} d u

Complete the square

1 + 2 u^{2} + 2 u : 2 (u + \frac{1}{2})^{2} + \frac{1}{2}

= - \int \frac{1}{2 ( u + \frac{1}{2} ) ^{2} + \frac{1}{2}} d u

Apply u-substitution

: \int \frac{2}{4 v ^{2} + 1} d v

= - \int \frac{2}{4 v ^{2} + 1} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 2 \cdot \int \frac{1}{4 v ^{2} + 1} d v

Apply Integral Substitution

= - 2 \cdot \int \frac{1}{2 ( w ^{2} + 1 )} d w

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 2 \cdot \frac{1}{2} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

= - 2 \cdot \frac{1}{2} arctan (w)

Substitute back

= - 2 \cdot \frac{1}{2} arctan (2 (e^{- x} + \frac{1}{2}))

Simplify

- 2 \cdot \frac{1}{2} arctan (2 (e^{- x} + \frac{1}{2})) : - arctan (2 e^{- x} + 1)

= - arctan (2 e^{- x} + 1)

Add a constant to the solution

= - arctan (2 e^{- x} + 1) + C

\int \frac{csc ( \frac{5}{x ^{3}} )}{x ^{4}} d x

f (x) = x^{2} + x, (2, - 3)

x \to 8 + lim (\frac{1}{( x - 8 )})

(7 x + 3 y) d x + (3 x - 8 y 3) d y = 0

\frac{k x + a}{b x + c}

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