What is the integral from 3 to 9 of 2/(11-4x) ?

The integral from 3 to 9 of 2/(11-4x) is -ln(5)

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integral from 3 to 9 of 2/(11-4x)

\int_{3}^{9} \frac{2}{11 - 4 x} d x

- ln (5)

Decimal

- 1.60943 \dots

Solution steps

\int_{3}^{9} \frac{2}{11 - 4 x} d x

Take the constant out:

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

= 2 \cdot \int_{3}^{9} \frac{1}{11 - 4 x} d x

Apply u-substitution

: \int_{- 1}^{- 25} - \frac{1}{4 u} d u

= 2 \cdot \int_{- 1}^{- 25} - \frac{1}{4 u} d u

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

= 2 (- \int_{- 25}^{- 1} - \frac{1}{4 u} d u)

Take the constant out:

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

= 2 (- (- \frac{1}{4} \cdot \int_{- 25}^{- 1} \frac{1}{u} d u))

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 (- (- \frac{1}{4} [ln ∣ u ∣]_{- 25}^{- 1}))

Simplify

2 (- (- \frac{1}{4} [ln ∣ u ∣]_{- 25}^{- 1})) : \frac{1}{2} [ln ∣ u ∣]_{- 25}^{- 1}

= \frac{1}{2} [ln ∣ u ∣]_{- 25}^{- 1}

Compute the boundaries:

- 2 ln (5)

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

Simplify

= - ln (5)

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What is the integral from 3 to 9 of 2/(11-4x) ?
The integral from 3 to 9 of 2/(11-4x) is -ln(5)