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

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

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

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

\frac{1}{3} (ln (2) - ln (5))

Decimal Notation

- 0.30543 \dots

Solution steps

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

Apply u-substitution

: \int_{- 2}^{- 5} - \frac{1}{3 u} d u

= \int_{- 2}^{- 5} - \frac{1}{3 u} d u

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

= - \int_{- 5}^{- 2} - \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_{- 5}^{- 2} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{3} [ln ∣ u ∣]_{- 5}^{- 2})

Simplify

= \frac{1}{3} [ln ∣ u ∣]_{- 5}^{- 2}

Compute the boundaries:

ln (2) - ln (5)

= \frac{1}{3} (ln (2) - ln (5))

\int - 3 cos (x) d x

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\frac{d y}{d x} = \frac{1}{x - x ^{3}}

\frac{\partial}{\partial x} (2 x - 3 y + 5 z)

\int 3^{2 x} cos^{2} (3^{x}) d x

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