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

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

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

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

\frac{1}{2} ln ∣ 2 a + 1 ∣

Solution steps

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

Apply u-substitution

: \int_{1}^{2 a + 1} \frac{1}{2 u} d u

= \int_{1}^{2 a + 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

= \frac{1}{2} \cdot \int_{1}^{2 a + 1} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{1}^{2 a + 1}

Compute the boundaries:

ln ∣ 2 a + 1 ∣

= \frac{1}{2} ln ∣ 2 a + 1 ∣

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