What is the integral from 1 to 2 of 1/(y+4) ?

The integral from 1 to 2 of 1/(y+4) is ln(6)-ln(5)

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integral from 1 to 2 of 1/(y+4)

\int_{1}^{2} \frac{1}{y + 4} d y

ln (6) - ln (5)

Solution steps

\int_{1}^{2} \frac{1}{y + 4} d y

Apply u-substitution

: \int_{5}^{6} \frac{1}{u} d u

= \int_{5}^{6} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{5}^{6}

Compute the boundaries:

ln (6) - ln (5)

= ln (6) - ln (5)

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What is the integral from 1 to 2 of 1/(y+4) ?
The integral from 1 to 2 of 1/(y+4) is ln(6)-ln(5)