What is the integral from-3 to 0 of 1/(9+2x) ?

The integral from-3 to 0 of 1/(9+2x) is 1/2 ln(3)

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integral from-3 to 0 of 1/(9+2x)

\int_{- 3}^{0} \frac{1}{9 + 2 x} d x

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

Decimal Notation

0.54930 \dots

Solution steps

\int_{- 3}^{0} \frac{1}{9 + 2 x} 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)

x \to 2 lim (\frac{5 x - 1}{x - 2})

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

\int arcsin (x + 2) d x

\frac{1}{s ^{3} - 2 s ^{2} + 3 s}

y = arcsinh (tan (x))

What is the integral from-3 to 0 of 1/(9+2x) ?
The integral from-3 to 0 of 1/(9+2x) is 1/2 ln(3)