What is the integral from 0 to 1 of (e^x)/(5-e^x) ?

The integral from 0 to 1 of (e^x)/(5-e^x) is 2ln(2)-ln(5-e)

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integral from 0 to 1 of (e^x)/(5-e^x)

\int_{0}^{1} \frac{e ^{x}}{5 - e ^{x}} d x

2 ln (2) - ln (5 - e)

Decimal

0.56136 \dots

Solution steps

\int_{0}^{1} \frac{e ^{x}}{5 - e ^{x}} d x

Apply u-substitution

: \int_{4}^{5 - e} - \frac{1}{u} d u

= \int_{4}^{5 - e} - \frac{1}{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{5 - e}^{4} - \frac{1}{u} d u

Take the constant out:

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

= - (- \int_{5 - e}^{4} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- [ln ∣ u ∣]_{5 - e}^{4})

Simplify

= [ln ∣ u ∣]_{5 - e}^{4}

Compute the boundaries:

2 ln (2) - ln (5 - e)

= 2 ln (2) - ln (5 - e)

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What is the integral from 0 to 1 of (e^x)/(5-e^x) ?
The integral from 0 to 1 of (e^x)/(5-e^x) is 2ln(2)-ln(5-e)