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

The integral of ((1+e^{2x})^3)/(e^{-2x)} is 1/8 (1+e^{2x})^4+C

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

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

\frac{1}{8} (1 + e^{2 x})^{4} + C

Solution steps

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

Simplify

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

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

Apply u-substitution

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

= \int \frac{u ^{3}}{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^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= \frac{1}{2} \cdot \frac{u ^{4}}{4}

Substitute back

u = 1 + e^{2 x}

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

Simplify

\frac{1}{2} \cdot \frac{( 1 + e ^{2 x} ) ^{4}}{4} : \frac{1}{8} (1 + e^{2 x})^{4}

= \frac{1}{8} (1 + e^{2 x})^{4}

Add a constant to the solution

= \frac{1}{8} (1 + e^{2 x})^{4} + C

y = 3^{x} + 1

\int (sin (x) - cos (2 x))^{2} d x

\int (7 + 7 tan^{2} (θ)) d θ

\frac{d}{d x} (\frac{x ^{4} - 3 x ^{3} - 6 x ^{2} - 3 x + 2}{x})

f (x) = \frac{x}{( 2 x - 1 ) ^{6}}, at x = 1

What is the integral of ((1+e^{2x})^3)/(e^{-2x)} ?
The integral of ((1+e^{2x})^3)/(e^{-2x)} is 1/8 (1+e^{2x})^4+C