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

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

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

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

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

Solution steps

\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

y^{^{''}} + 2 y^{^{'}} + 2 y = cos (2 x)

\frac{d}{d x} (\frac{x}{y ^{2}})

\frac{\partial}{\partial x} (sin (- 2 x) cos (4 y))

\frac{d}{d x} (x^{2} + a \cdot x + b)

\frac{d y}{d x} = x^{2} (1 + y)

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