What is the integral from 2 to 4 of 1/(3t+4) ?

The integral from 2 to 4 of 1/(3t+4) is 1/3 ln(8/5)

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integral from 2 to 4 of 1/(3t+4)

\int_{2}^{4} \frac{1}{3 t + 4} d t

\frac{1}{3} ln (\frac{8}{5})

Decimal Notation

0.15666 \dots

Solution steps

\int_{2}^{4} \frac{1}{3 t + 4} d t

Apply u-substitution

: \int_{10}^{16} \frac{1}{3 u} d u

= \int_{10}^{16} \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{10}^{16} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{10}^{16}

Compute the boundaries:

4 ln (2) - ln (10)

= \frac{1}{3} (4 ln (2) - ln (10))

Simplify

= \frac{1}{3} ln (\frac{8}{5})

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What is the integral from 2 to 4 of 1/(3t+4) ?
The integral from 2 to 4 of 1/(3t+4) is 1/3 ln(8/5)