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

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

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

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

2

Solution steps

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

Apply u-substitution

: \int_{\frac{1}{e ^{2}} + 1}^{e^{2} + 1} \frac{1}{u} d u

= \int_{\frac{1}{e ^{2}} + 1}^{e^{2} + 1} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{\frac{1}{e ^{2}} + 1}^{e^{2} + 1}

Compute the boundaries:

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

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

Simplify

= 2

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