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

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

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

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

e^{x} + 1 - ln ∣ e^{x} + 1 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{u - 1}{u} d u

= \int \frac{u - 1}{u} d u

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= \int 1 - \frac{1}{u} d u

Apply the Sum Rule:

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

= \int 1 d u - \int \frac{1}{u} d u

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= u - ln ∣ u ∣

Substitute back

u = e^{x} + 1

= e^{x} + 1 - ln ∣ e^{x} + 1 ∣

Add a constant to the solution

= e^{x} + 1 - ln ∣ e^{x} + 1 ∣ + C

x \to 0 lim (\frac{( x + 2 ) ^{2} - 8}{x})

x^{2} + 8 x - 3, - x^{2} + 4 x + 3

n = 1 \sum \infty \frac{5 ^{n}}{9}

f (x) = \frac{sin ( 4 x )}{sin ( 5 x )}

y = arctan (3 x^{2})

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