What is the integral from 0 to 4 of 1/(2x+1) ?

The integral from 0 to 4 of 1/(2x+1) is ln(3)

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integral from 0 to 4 of 1/(2x+1)

\int_{0}^{4} \frac{1}{2 x + 1} d x

ln (3)

Decimal

1.09861 \dots

Solution steps

\int_{0}^{4} \frac{1}{2 x + 1} d x

Apply u-substitution

= \int_{1}^{9} \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}^{9} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{1}^{9}

Compute the boundaries:

2 ln (3)

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

Simplify

= ln (3)

What is the integral from 0 to 4 of 1/(2x+1) ?
The integral from 0 to 4 of 1/(2x+1) is ln(3)