What is the integral from 0 to 1 of (e^z+9)/(e^z+9z) ?

The integral from 0 to 1 of (e^z+9)/(e^z+9z) is ln(e+9)

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integral from 0 to 1 of (e^z+9)/(e^z+9z)

\int_{0}^{1} \frac{e ^{z} + 9}{e ^{z} + 9 z} d z

ln (e + 9)

Solution steps

\int_{0}^{1} \frac{e ^{z} + 9}{e ^{z} + 9 z} d z

Apply u-substitution

: \int_{1}^{e + 9} \frac{1}{u} d u

= \int_{1}^{e + 9} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{e + 9}

Compute the boundaries:

ln (e + 9)

= ln (e + 9)

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What is the integral from 0 to 1 of (e^z+9)/(e^z+9z) ?
The integral from 0 to 1 of (e^z+9)/(e^z+9z) is ln(e+9)