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

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

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

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

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

Decimal Notation

0.54930 \dots

Solution steps

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

Apply u-substitution

: \int_{3}^{9} \frac{1}{2 u} d u

= \int_{3}^{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_{3}^{9} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{3}^{9}

Compute the boundaries:

ln (3)

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

\frac{d}{d x} (x - ln (x e^{x}))

\frac{d}{d x} (2 - \frac{2}{x})

\int \frac{1}{x ^{\frac{2}{5}}} d x

\int_{0}^{2} x^{2} d x

\int e^{x^{2}} x^{2} d x

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