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

The integral of (e^{2x})/((e^x+1)^2) is ln|e^x+1|+1/(e^x+1)+C

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

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

ln ∣ e^{x} + 1 ∣ + \frac{1}{e ^{x} + 1} + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{2 v}{( v + 1 ) ^{2}} d v

= \frac{1}{2} \cdot \int \frac{2 v}{( v + 1 ) ^{2}} d v

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int \frac{v}{( v + 1 ) ^{2}} d v

Apply u-substitution

: \int \frac{w - 1}{w ^{2}} d w

= \frac{1}{2} \cdot 2 \cdot \int \frac{w - 1}{w ^{2}} d w

Expand

\frac{w - 1}{w ^{2}} : \frac{1}{w} - \frac{1}{w ^{2}}

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{w} - \frac{1}{w ^{2}} d w

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} \cdot 2 (\int \frac{1}{w} d w - \int \frac{1}{w ^{2}} d w)

\int \frac{1}{w} d w = ln ∣ w ∣

\int \frac{1}{w ^{2}} d w = - \frac{1}{w}

= \frac{1}{2} \cdot 2 (ln ∣ w ∣ - (- \frac{1}{w}))

Substitute back

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

Simplify

\frac{1}{2} \cdot 2 (ln e^{2 x} + 1 - (- \frac{1}{e ^{2 x} + 1})) : ln ∣ e^{x} + 1 ∣ + \frac{1}{e ^{x} + 1}

= ln ∣ e^{x} + 1 ∣ + \frac{1}{e ^{x} + 1}

Add a constant to the solution

= ln ∣ e^{x} + 1 ∣ + \frac{1}{e ^{x} + 1} + C

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What is the integral of (e^{2x})/((e^x+1)^2) ?
The integral of (e^{2x})/((e^x+1)^2) is ln|e^x+1|+1/(e^x+1)+C