What is the integral of (sqrt(1+e^{-2x)})/(e^{-3x)} ?

The integral of (sqrt(1+e^{-2x)})/(e^{-3x)} is ((1+e^{-2x})^{3/2})/(3e^{-3x)}+C

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integral of (sqrt(1+e^{-2x)})/(e^{-3x)}

\int \frac{1 + e ^{- 2 x}}{e ^{- 3 x}} d x

\frac{( 1 + e ^{- 2 x} ) ^{\frac{3}{2}}}{3 e ^{- 3 x}} + C

Solution steps

\int \frac{1 + e ^{- 2 x}}{e ^{- 3 x}} d x

Apply u-substitution

: \int - \frac{u ^{2}}{( u ^{2} - 1 ) ^{\frac{5}{2}}} d u

= \int - \frac{u ^{2}}{( u ^{2} - 1 ) ^{\frac{5}{2}}} d u

Take the constant out:

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

= - \int \frac{u ^{2}}{( u ^{2} - 1 ) ^{\frac{5}{2}}} d u

Apply Trigonometric Substitution

: \int cot^{4} (v) sec^{3} (v) d v

= - \int cot^{4} (v) sec^{3} (v) d v

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{1}{w ^{4}} d w

= - \int \frac{1}{w ^{4}} d w

Apply the Power Rule

: - \frac{1}{3 w ^{3}}

= - (- \frac{1}{3 w ^{3}})

Substitute back

= - (- \frac{1}{3 sin ^{3} ( arcsec ( 1 + e ^{- 2 x} ) )})

Simplify

- (- \frac{1}{3 sin ^{3} ( arcsec ( 1 + e ^{- 2 x} ) )}) : \frac{( 1 + e ^{- 2 x} ) ^{\frac{3}{2}}}{3 e ^{- 3 x}}

= \frac{( 1 + e ^{- 2 x} ) ^{\frac{3}{2}}}{3 e ^{- 3 x}}

Add a constant to the solution

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

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What is the integral of (sqrt(1+e^{-2x)})/(e^{-3x)} ?
The integral of (sqrt(1+e^{-2x)})/(e^{-3x)} is ((1+e^{-2x})^{3/2})/(3e^{-3x)}+C