What is the integral of (e^{2x})/((1+e^{2x))^4} ?

The integral of (e^{2x})/((1+e^{2x))^4} is -1/(6(1+e^{2x))^3}+C

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integral of (e^{2x})/((1+e^{2x))^4}

\int \frac{e ^{2 x}}{( 1 + e ^{2 x} ) ^{4}} d x

- \frac{1}{6 ( 1 + e ^{2 x} ) ^{3}} + C

Solution steps

\int \frac{e ^{2 x}}{( 1 + e ^{2 x} ) ^{4}} d x

Apply u-substitution

: \int \frac{e ^{u}}{2 ( 1 + e ^{u} ) ^{4}} d u

= \int \frac{e ^{u}}{2 ( 1 + e ^{u} ) ^{4}} 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 \frac{e ^{u}}{( 1 + e ^{u} ) ^{4}} d u

Apply u-substitution

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

= \frac{1}{2} \cdot \int \frac{1}{v ^{4}} d v

Apply the Power Rule

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

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

Substitute back

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

Simplify

\frac{1}{2} (- \frac{1}{3 ( 1 + e ^{2 x} ) ^{3}}) : - \frac{1}{6 ( 1 + e ^{2 x} ) ^{3}}

= - \frac{1}{6 ( 1 + e ^{2 x} ) ^{3}}

Add a constant to the solution

= - \frac{1}{6 ( 1 + e ^{2 x} ) ^{3}} + C

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What is the integral of (e^{2x})/((1+e^{2x))^4} ?
The integral of (e^{2x})/((1+e^{2x))^4} is -1/(6(1+e^{2x))^3}+C