What is the integral from 0 to 1 of 1/(3(1+y)) ?

The integral from 0 to 1 of 1/(3(1+y)) is 1/3 ln(2)

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integral from 0 to 1 of 1/(3(1+y))

\int_{0}^{1} \frac{1}{3 ( 1 + y )} d y

\frac{1}{3} ln (2)

Decimal

0.23104 \dots

Solution steps

\int_{0}^{1} \frac{1}{3 ( 1 + y )} d y

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{1} \frac{1}{1 + y} d y

Apply u-substitution

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

= \frac{1}{3} \cdot \int_{1}^{2} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= \frac{1}{3} ln (2)

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What is the integral from 0 to 1 of 1/(3(1+y)) ?
The integral from 0 to 1 of 1/(3(1+y)) is 1/3 ln(2)