What is the integral from-3 to-2 of ((x+2))/(x^2+4x) ?

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

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integral from-3 to-2 of ((x+2))/(x^2+4x)

\int_{- 3}^{- 2} \frac{( x + 2 )}{x ^{2} + 4 x} d x

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

Solution steps

\int_{- 3}^{- 2} \frac{x + 2}{x ^{2} + 4 x} d x

Apply u-substitution

: \int_{- 3}^{- 4} \frac{1}{2 u} d u

= \int_{- 3}^{- 4} \frac{1}{2 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 4}^{- 3} \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_{- 4}^{- 3} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} [ln ∣ u ∣]_{- 4}^{- 3}

Compute the boundaries:

ln (3) - 2 ln (2)

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

Simplify

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

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