What is the integral from 2 to 4 of 6/(2x-9) ?

The integral from 2 to 4 of 6/(2x-9) is -3ln(5)

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integral from 2 to 4 of 6/(2x-9)

\int_{2}^{4} \frac{6}{2 x - 9} d x

- 3 ln (5)

Decimal Notation

- 4.82831 \dots

Solution steps

\int_{2}^{4} \frac{6}{2 x - 9} d x

Take the constant out:

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

= 6 \cdot \int_{2}^{4} \frac{1}{2 x - 9} d x

Apply u-substitution

: \int_{- 5}^{- 1} \frac{1}{2 u} d u

= 6 \cdot \int_{- 5}^{- 1} \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_{- 5}^{- 1} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 \cdot \frac{1}{2} [ln ∣ u ∣]_{- 5}^{- 1}

Simplify

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

= 3 [ln ∣ u ∣]_{- 5}^{- 1}

Compute the boundaries:

- ln (5)

= 3 (- ln (5))

Simplify

= - 3 ln (5)

x \to 8 lim (x + 2)

f (x) = ln (x^{3} - 7), at x = 2

\int \frac{1}{x x + x ^{2}} d x

\int - 5 sin (x) d x

\int_{0}^{L} sin^{2} (\frac{nπ x}{L}) d x

What is the integral from 2 to 4 of 6/(2x-9) ?
The integral from 2 to 4 of 6/(2x-9) is -3ln(5)