What is the integral of 1/(e^xsqrt(4e^{-x)+e^{-2x)}} ?

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

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integral of 1/(e^xsqrt(4e^{-x)+e^{-2x)}}

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

- ln 4 e^{x} + 1 + 1 + ln 4 e^{x} + 1 - 1 + C

Solution steps

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

Multiply by the conjugate of

4 e^{- x} + e^{- 2 x} : \frac{16 e ^{- 2 x} - e ^{- 4 x}}{4 e ^{- x} - e ^{- 2 x}}

= \int \frac{1}{e ^{x} \frac{16 e ^{- 2 x} - e ^{- 4 x}}{4 e ^{- x} - e ^{- 2 x}}} d x

Simplify

\frac{1}{e ^{x} \frac{16 e ^{- 2 x} - e ^{- 4 x}}{4 e ^{- x} - e ^{- 2 x}}} : \frac{1}{4 e ^{x} + 1}

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

Apply u-substitution

: \int \frac{1}{u 4 u + 1} d u

= \int \frac{1}{u 4 u + 1} d u

Apply u-substitution

: \int \frac{2}{v ^{2} - 1} d v

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

Factor

v^{2} - 1 : - (- v^{2} + 1)

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

Take the constant out:

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

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

Use the common integral:

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

= 2 (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

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

Simplify

2 (- (\frac{ln 4 e ^{x} + 1 + 1}{2} - \frac{ln 4 e ^{x} + 1 - 1}{2})) : - ln 4 e^{x} + 1 + 1 + ln 4 e^{x} + 1 - 1

= - ln 4 e^{x} + 1 + 1 + ln 4 e^{x} + 1 - 1

Add a constant to the solution

= - ln 4 e^{x} + 1 + 1 + ln 4 e^{x} + 1 - 1 + C

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