What is the integral from 2 to 4 of 1/(5-3x) ?

The integral from 2 to 4 of 1/(5-3x) is -1/3 ln(7)

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integral from 2 to 4 of 1/(5-3x)

\int_{2}^{4} \frac{1}{5 - 3 x} d x

- \frac{1}{3} ln (7)

Decimal Notation

- 0.64863 \dots

Solution steps

\int_{2}^{4} \frac{1}{5 - 3 x} d x

Apply u-substitution

: \int_{- 1}^{- 7} - \frac{1}{3 u} d u

= \int_{- 1}^{- 7} - \frac{1}{3 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 7}^{- 1} - \frac{1}{3 u} d u

Take the constant out:

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

= - (- \frac{1}{3} \cdot \int_{- 7}^{- 1} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{3} [ln ∣ u ∣]_{- 7}^{- 1})

Simplify

= \frac{1}{3} [ln ∣ u ∣]_{- 7}^{- 1}

Compute the boundaries:

- ln (7)

= \frac{1}{3} (- ln (7))

Simplify

= - \frac{1}{3} ln (7)

\int_{6}^{8} \frac{2 x}{x - 5} d x

\frac{d}{d x} (\frac{4}{x - 7})

f (x) = (4 x - 2)^{\frac{1}{5}}, at x = 0

\int 2 x^{3} + x d x

f (x) = 2 x^{2} + 5 x - 3, at x = 1

What is the integral from 2 to 4 of 1/(5-3x) ?
The integral from 2 to 4 of 1/(5-3x) is -1/3 ln(7)