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

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

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

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

ln (2)

Decimal

0.69314 \dots

Solution steps

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

Apply u-substitution

: \int_{4}^{1} - \frac{1}{2 u} d u

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

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

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

Simplify

= \frac{1}{2} [ln ∣ u ∣]_{1}^{4}

Compute the boundaries:

2 ln (2)

= \frac{1}{2} \cdot 2 ln (2)

Simplify

= ln (2)

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