What is the integral from 3 to 6 of 6/(2x-3) ?

The integral from 3 to 6 of 6/(2x-3) is 3ln(3)

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integral from 3 to 6 of 6/(2x-3)

\int_{3}^{6} \frac{6}{2 x - 3} d x

3 ln (3)

Decimal Notation

3.29583 \dots

Solution steps

\int_{3}^{6} \frac{6}{2 x - 3} d x

Take the constant out:

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

= 6 \cdot \int_{3}^{6} \frac{1}{2 x - 3} d x

Apply u-substitution

: \int_{3}^{9} \frac{1}{2 u} d u

= 6 \cdot \int_{3}^{9} \frac{1}{2 u} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int_{3}^{9} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 \cdot \frac{1}{2} [ln ∣ u ∣]_{3}^{9}

Simplify

6 \cdot \frac{1}{2} [ln ∣ u ∣]_{3}^{9} : 3 [ln ∣ u ∣]_{3}^{9}

= 3 [ln ∣ u ∣]_{3}^{9}

Compute the boundaries:

ln (3)

= 3 ln (3)

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