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

The integral from 0 to 1 of (1+e^x)/(e^x+x) is ln(e+1)

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

\int_{0}^{1} \frac{1 + e ^{x}}{e ^{x} + x} d x

ln (e + 1)

Solution steps

\int_{0}^{1} \frac{1 + e ^{x}}{e ^{x} + x} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{e + 1}

Compute the boundaries:

ln (e + 1)

= ln (e + 1)

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