What is the integral from 2 to 9 of 4/(8x-9) ?

The integral from 2 to 9 of 4/(8x-9) is ln(3)

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integral from 2 to 9 of 4/(8x-9)

\int_{2}^{9} \frac{4}{8 x - 9} d x

ln (3)

Decimal

1.09861 \dots

Solution steps

\int_{2}^{9} \frac{4}{8 x - 9} d x

Take the constant out:

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

= 4 \cdot \int_{2}^{9} \frac{1}{8 x - 9} d x

Apply u-substitution

= 4 \cdot \int_{7}^{63} \frac{1}{8 u} d u

Take the constant out:

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

= 4 \cdot \frac{1}{8} \cdot \int_{7}^{63} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{8} [ln ∣ u ∣]_{7}^{63}

Simplify

4 \cdot \frac{1}{8} [ln ∣ u ∣]_{7}^{63} : \frac{1}{2} [ln ∣ u ∣]_{7}^{63}

= \frac{1}{2} [ln ∣ u ∣]_{7}^{63}

Compute the boundaries:

ln (63) - ln (7)

= \frac{1}{2} (ln (63) - ln (7))

Simplify

= ln (3)

What is the integral from 2 to 9 of 4/(8x-9) ?
The integral from 2 to 9 of 4/(8x-9) is ln(3)