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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int \frac{cot ( v )}{4 cos ( v ) sin ( v )} d v

= - \int \frac{cot ( v )}{4 cos ( v ) sin ( v )} d v

Take the constant out:

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

= - \frac{1}{4} \cdot \int \frac{cot ( v )}{cos ( v ) sin ( v )} d v

Simplify

\frac{cot ( v )}{cos ( v ) sin ( v )} : csc^{2} (v)

= - \frac{1}{4} \cdot \int csc^{2} (v) d v

Use the common integral:

\int csc^{2} (v) d v = - cot (v)

= - \frac{1}{4} (- cot (v))

Substitute back

= - \frac{1}{4} (- cot (arcsin (\frac{1}{2} e^{- x})))

Simplify

- \frac{1}{4} (- cot (arcsin (\frac{1}{2} e^{- x}))) : \frac{1}{4} e^{x} 4 - e^{- 2 x}

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

Add a constant to the solution

= \frac{1}{4} e^{x} 4 - e^{- 2 x} + C

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