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

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

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

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

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

Solution steps

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

Simplify

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

= \int e^{2 x} 1 + e^{2 x} d x

Apply u-substitution

: \int \frac{u}{2} d u

= \int \frac{u}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + e^{2 x}

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

Simplify

\frac{1}{2} \cdot \frac{2}{3} (1 + e^{2 x})^{\frac{3}{2}} : \frac{1}{3} (1 + e^{2 x})^{\frac{3}{2}}

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

Add a constant to the solution

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

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