What is the integral from 0 to 1 of 1/(1+2x) ?

The integral from 0 to 1 of 1/(1+2x) is 1/2 ln(3)

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integral from 0 to 1 of 1/(1+2x)

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

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

Solution steps

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

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What is the integral from 0 to 1 of 1/(1+2x) ?
The integral from 0 to 1 of 1/(1+2x) is 1/2 ln(3)