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

The integral from 0 to 2 of 1/(2x+3) is 1/2 (ln(7)-ln(3))

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

\int_{0}^{2} \frac{1}{2 x + 3} d x

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

Decimal

0.42364 \dots

Solution steps

\int_{0}^{2} \frac{1}{2 x + 3} 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_{1}^{x} (2 - x) d x

\int_{1}^{4} x x d x

\int_{- 2}^{2} 3 x d x

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

\int_{2}^{3} 12 (x^{2} - 4)^{5} x d x

What is the integral from 0 to 2 of 1/(2x+3) ?
The integral from 0 to 2 of 1/(2x+3) is 1/2 (ln(7)-ln(3))