What is the integral from 0 to 1 of 1/(6x+2) ?

The integral from 0 to 1 of 1/(6x+2) is 1/3 ln(2)

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integral from 0 to 1 of 1/(6x+2)

\int_{0}^{1} \frac{1}{6 x + 2} d x

\frac{1}{3} ln (2)

Decimal Notation

0.23104 \dots

Solution steps

\int_{0}^{1} \frac{1}{6 x + 2} d x

Apply u-substitution

: \int_{2}^{8} \frac{1}{6 u} d u

= \int_{2}^{8} \frac{1}{6 u} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int_{2}^{8} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} [ln ∣ u ∣]_{2}^{8}

Compute the boundaries:

2 ln (2)

= \frac{1}{6} \cdot 2 ln (2)

Simplify

= \frac{1}{3} ln (2)

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What is the integral from 0 to 1 of 1/(6x+2) ?
The integral from 0 to 1 of 1/(6x+2) is 1/3 ln(2)