What is the integral from 5 to 10 of 1/(t+1) ?

The integral from 5 to 10 of 1/(t+1) is ln(11)-ln(6)

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integral from 5 to 10 of 1/(t+1)

\int_{5}^{10} \frac{1}{t + 1} d t

ln (11) - ln (6)

Decimal Notation

0.60613 \dots

Solution steps

\int_{5}^{10} \frac{1}{t + 1} d t

Apply u-substitution

: \int_{6}^{11} \frac{1}{u} d u

= \int_{6}^{11} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{6}^{11}

Compute the boundaries:

ln (11) - ln (6)

= ln (11) - ln (6)

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\int_{1}^{3} (2 x^{3} - 3 x^{- 2}) d x

What is the integral from 5 to 10 of 1/(t+1) ?
The integral from 5 to 10 of 1/(t+1) is ln(11)-ln(6)