What is the integral from-1 to 3 of (e^{2x})/(e^x+1) ?

The integral from-1 to 3 of (e^{2x})/(e^x+1) is e^3-1/e-ln((e^4+e)/(1+e))

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integral from-1 to 3 of (e^{2x})/(e^x+1)

\int_{- 1}^{3} \frac{e ^{2 x}}{e ^{x} + 1} d x

e^{3} - \frac{1}{e} - ln (\frac{e ^{4} + e}{1 + e})

Decimal Notation

16.98233 \dots

Solution steps

\int_{- 1}^{3} \frac{e ^{2 x}}{e ^{x} + 1} d x

Apply u-substitution

: \int_{\frac{1}{e} + 1}^{e^{3} + 1} \frac{u - 1}{u} d u

= \int_{\frac{1}{e} + 1}^{e^{3} + 1} \frac{u - 1}{u} d u

Expand

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

= \int_{\frac{1}{e} + 1}^{e^{3} + 1} 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_{\frac{1}{e} + 1}^{e^{3} + 1} 1 d u - \int_{\frac{1}{e} + 1}^{e^{3} + 1} \frac{1}{u} d u

\int_{\frac{1}{e} + 1}^{e^{3} + 1} 1 d u = e^{3} - \frac{1}{e}

\int_{\frac{1}{e} + 1}^{e^{3} + 1} \frac{1}{u} d u = ln (\frac{e ^{4} + e}{1 + e})

= e^{3} - \frac{1}{e} - ln (\frac{e ^{4} + e}{1 + e})

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