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

The integral from-1 to 1 of (3e^x)/(1+e^x) is 3

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

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

3

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 [ln ∣ u ∣]_{1 + \frac{1}{e}}^{1 + e}

Compute the boundaries:

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

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

Simplify

= 3

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