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

The integral from-1 to 1 of (e^{2x})/(1+e^x) is e-1/e-1

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

\int_{- 1}^{1} \frac{e ^{2 x}}{1 + e ^{x}} d x

e - \frac{1}{e} - 1

Solution steps

\int_{- 1}^{1} \frac{e ^{2 x}}{1 + e ^{x}} d x

Apply u-substitution

: \int_{1 + \frac{1}{e}}^{1 + e} \frac{u - 1}{u} d u

= \int_{1 + \frac{1}{e}}^{1 + e} \frac{u - 1}{u} d u

Expand

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

= \int_{1 + \frac{1}{e}}^{1 + e} 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 + \frac{1}{e}}^{1 + e} 1 d u - \int_{1 + \frac{1}{e}}^{1 + e} \frac{1}{u} d u

\int_{1 + \frac{1}{e}}^{1 + e} 1 d u = e - \frac{1}{e}

\int_{1 + \frac{1}{e}}^{1 + e} \frac{1}{u} d u = 1

= e - \frac{1}{e} - 1

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

f (x) = x^{2}, (- 0.8, 0.64)

\int \frac{5 x}{( x ^{2} + 6 ) ^{2}} d x

\int 0 e^{x} d x

y^{^{''}} sin (x) cos (x) - 2 y^{^{'}} = 0

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