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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Decimal Notation

0.14384 \dots

Solution steps

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

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)

\int_{- 3}^{1} \frac{1}{( 2 x + 6 ) ^{\frac{2}{3}}} d x

\int_{- \infty}^{0} 2 e^{- 3 x} d x

\int_{- 6}^{- 2} (- x + 1) d x

\int_{0}^{π} sin (x) e^{- s x} d x

\int_{0}^{t} e^{- (t - x)} d x

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