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

The integral from 2 to 3 of 2/(3x-1) is 2ln(2)-2/3 ln(5)

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integral from 2 to 3 of 2/(3x-1)

\int_{2}^{3} \frac{2}{3 x - 1} d x

2 ln (2) - \frac{2}{3} ln (5)

Decimal Notation

0.31333 \dots

Solution steps

\int_{2}^{3} \frac{2}{3 x - 1} d x

Take the constant out:

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

= 2 \cdot \int_{2}^{3} \frac{1}{3 x - 1} d x

Apply u-substitution

: \int_{5}^{8} \frac{1}{3 u} d u

= 2 \cdot \int_{5}^{8} \frac{1}{3 u} d u

Take the constant out:

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

= 2 \cdot \frac{1}{3} \cdot \int_{5}^{8} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{3} [ln ∣ u ∣]_{5}^{8}

Simplify

2 \cdot \frac{1}{3} [ln ∣ u ∣]_{5}^{8} : \frac{2}{3} [ln ∣ u ∣]_{5}^{8}

= \frac{2}{3} [ln ∣ u ∣]_{5}^{8}

Compute the boundaries:

3 ln (2) - ln (5)

= \frac{2}{3} (3 ln (2) - ln (5))

Simplify

= 2 ln (2) - \frac{2}{3} ln (5)

\int_{- 5}^{2} x^{2} + 3 x - 10 d x

\int_{1}^{4} 5 x^{2} + x d x

\int_{0}^{4} e^{- 2 x} d x

\int_{0}^{π} x^{2} sin (x) cos (x) d x

\int_{- 2}^{2} x (x^{2} - 5) d x

What is the integral from 2 to 3 of 2/(3x-1) ?
The integral from 2 to 3 of 2/(3x-1) is 2ln(2)-2/3 ln(5)