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

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

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

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

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

Decimal Notation

0.54930 \dots

Solution steps

\int_{1}^{2} \frac{1}{2 x - 1} 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 27 - x^{2} d x

x \to \frac{π}{2} lim (e^{t a n (x)})

y = \frac{4 x}{x ^{2} + 1}

\int 38 d x

sin (t + 2 π)

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