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

The integral from 0 to 2 of 1/(3x+1) is 1/3 ln(7)

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integral from 0 to 2 of 1/(3x+1)

\int_{0}^{2} \frac{1}{3 x + 1} d x

\frac{1}{3} ln (7)

Decimal

0.64863 \dots

Solution steps

\int_{0}^{2} \frac{1}{3 x + 1} d x

Apply u-substitution

= \int_{1}^{7} \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_{1}^{7} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{1}^{7}

Compute the boundaries:

ln (7)

= \frac{1}{3} ln (7)

What is the integral from 0 to 2 of 1/(3x+1) ?
The integral from 0 to 2 of 1/(3x+1) is 1/3 ln(7)