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

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

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

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

- \frac{3}{2} ln (7)

Decimal

- 2.91886 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 3 \cdot \int_{- 1}^{- 7} - \frac{1}{2 u} d u

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

= 3 (- \int_{- 7}^{- 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

= 3 (- (- \frac{1}{2} \cdot \int_{- 7}^{- 1} \frac{1}{u} d u))

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 (- (- \frac{1}{2} [ln ∣ u ∣]_{- 7}^{- 1}))

Simplify

3 (- (- \frac{1}{2} [ln ∣ u ∣]_{- 7}^{- 1})) : \frac{3}{2} [ln ∣ u ∣]_{- 7}^{- 1}

= \frac{3}{2} [ln ∣ u ∣]_{- 7}^{- 1}

Compute the boundaries:

- ln (7)

= \frac{3}{2} (- ln (7))

Simplify

= - \frac{3}{2} ln (7)

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