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

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

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

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

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

Decimal

0.54930 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{1}^{3}

Compute the boundaries:

ln (3)

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

\int_{- 2}^{2} 16 - x^{4} d x

\int_{0}^{2} (2 x + 1)^{3} d x

\int_{0}^{b} \frac{x}{( 2 x ^{2} + 1 ) ^{2}} d x

\int_{0}^{3} 4 x e^{2 x^{2}} d x

\int_{0}^{3} \frac{x}{25 - x ^{2}} d x

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