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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int - v d v

= \frac{1}{2} \cdot \int - v d v

Take the constant out:

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

= \frac{1}{2} (- \int v d v)

Apply the Power Rule

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

= \frac{1}{2} (- \frac{2}{3} v^{\frac{3}{2}})

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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