What is the integral from 0 to 9 of 1/(3+t) ?

The integral from 0 to 9 of 1/(3+t) is 2ln(2)

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integral from 0 to 9 of 1/(3+t)

\int_{0}^{9} \frac{1}{3 + t} d t

2 ln (2)

Decimal Notation

1.38629 \dots

Solution steps

\int_{0}^{9} \frac{1}{3 + t} d t

Apply u-substitution

: \int_{3}^{12} \frac{1}{u} d u

= \int_{3}^{12} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{3}^{12}

Compute the boundaries:

ln (12) - ln (3)

= ln (12) - ln (3)

Simplify

= 2 ln (2)

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What is the integral from 0 to 9 of 1/(3+t) ?
The integral from 0 to 9 of 1/(3+t) is 2ln(2)