What is the integral of 1/(e^x(3+e^{-x))} ?

The integral of 1/(e^x(3+e^{-x))} is -ln|3+e^{-x}|+C

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integral of 1/(e^x(3+e^{-x))}

\int \frac{1}{e ^{x} ( 3 + e ^{- x} )} d x

- ln 3 + e^{- x} + C

Solution steps

\int \frac{1}{e ^{x} ( 3 + e ^{- x} )} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{e ^{x} ( 3 + e ^{- x} )} = e^{- x} \frac{1}{( 3 + e ^{- x} )}

= \int e^{- x} \frac{1}{3 + e ^{- x}} d x

Simplify

e^{- x} \frac{1}{3 + e ^{- x}} : \frac{e ^{- x}}{3 + e ^{- x}}

= \int \frac{e ^{- x}}{3 + e ^{- x}} d x

Apply u-substitution

: \int - \frac{1}{u} d u

= \int - \frac{1}{u} d u

Take the constant out:

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

= - \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = 3 + e^{- x}

= - ln 3 + e^{- x}

Add a constant to the solution

= - ln 3 + e^{- x} + C

\frac{d}{d x} (e^{a x} sin (b x))

x^{2} \frac{d y}{d x} = 1 - x^{2} + y^{2} - x^{2} y^{2}

\int (x - a)^{2} d x

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

\frac{x - 13}{x - 13}

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