What is the integral from 1 to 4 of 9/(2x-1) ?

The integral from 1 to 4 of 9/(2x-1) is 9/2 ln(7)

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integral from 1 to 4 of 9/(2x-1)

\int_{1}^{4} \frac{9}{2 x - 1} d x

\frac{9}{2} ln (7)

Decimal

8.75659 \dots

Solution steps

\int_{1}^{4} \frac{9}{2 x - 1} d x

Take the constant out:

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

= 9 \cdot \int_{1}^{4} \frac{1}{2 x - 1} d x

Apply u-substitution

= 9 \cdot \int_{1}^{7} \frac{1}{2 u} d u

Take the constant out:

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

= 9 \cdot \frac{1}{2} \cdot \int_{1}^{7} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 9 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{7}

Simplify

9 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{7} : \frac{9}{2} [ln ∣ u ∣]_{1}^{7}

= \frac{9}{2} [ln ∣ u ∣]_{1}^{7}

Compute the boundaries:

ln (7)

= \frac{9}{2} ln (7)

What is the integral from 1 to 4 of 9/(2x-1) ?
The integral from 1 to 4 of 9/(2x-1) is 9/2 ln(7)