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

The integral from 1 to 3 of 1/(2x+1) is 1/2 (ln(7)-ln(3))

Popular >

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

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

\frac{1}{2} (ln (7) - ln (3))

Decimal Notation

0.42364 \dots

Solution steps

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

Apply u-substitution

: \int_{3}^{7} \frac{1}{2 u} d u

= \int_{3}^{7} \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_{3}^{7} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{3}^{7}

Compute the boundaries:

ln (7) - ln (3)

= \frac{1}{2} (ln (7) - ln (3))

\int_{4}^{25} x d x

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

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

\int_{0}^{\frac{π}{2}} (6 + 3 cos (x)) d x

\int_{x}^{1} 6 (1 - y) d y

What is the integral from 1 to 3 of 1/(2x+1) ?
The integral from 1 to 3 of 1/(2x+1) is 1/2 (ln(7)-ln(3))