What is the integral from 0 to 2 of 2/(y^2+y-20) ?

The integral from 0 to 2 of 2/(y^2+y-20) is -(2ln(14/5))/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 2 of 2/(y^2+y-20)

\int_{0}^{2} \frac{2}{y ^{2} + y - 20} d y

- \frac{2 ln ( \frac{14}{5} )}{9}

Decimal

- 0.22880 \dots

Solution steps

\int_{0}^{2} \frac{2}{y ^{2} + y - 20} d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{2} \frac{1}{y ^{2} + y - 20} d y

Complete the square

y^{2} + y - 20 : (y + \frac{1}{2})^{2} - \frac{81}{4}

= 2 \cdot \int_{0}^{2} \frac{1}{( y + \frac{1}{2} ) ^{2} - \frac{81}{4}} d y

Apply u-substitution

= 2 \cdot \int_{\frac{1}{2}}^{\frac{5}{2}} \frac{4}{4 u ^{2} - 81} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot 4 \cdot \int_{\frac{1}{2}}^{\frac{5}{2}} \frac{1}{4 u ^{2} - 81} d u

Apply Integral Substitution

= 2 \cdot 4 \cdot \int_{\frac{1}{9}}^{\frac{5}{9}} \frac{1}{18 ( v ^{2} - 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot 4 \cdot \frac{1}{18} \cdot \int_{\frac{1}{9}}^{\frac{5}{9}} \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

= 2 \cdot 4 \cdot \frac{1}{18} \cdot \int_{\frac{1}{9}}^{\frac{5}{9}} \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot 4 \cdot \frac{1}{18} (- \int_{\frac{1}{9}}^{\frac{5}{9}} \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 2 \cdot 4 \cdot \frac{1}{18} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{1}{9}}^{\frac{5}{9}})

Simplify

2 \cdot 4 \cdot \frac{1}{18} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{1}{9}}^{\frac{5}{9}}) : - \frac{4}{9} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{\frac{1}{9}}^{\frac{5}{9}}

= - \frac{4}{9} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{\frac{1}{9}}^{\frac{5}{9}}

Compute the boundaries:

\frac{ln ( \frac{14}{9} ) - ln ( \frac{4}{9} ) - ln ( \frac{10}{9} ) + ln ( \frac{8}{9} )}{2}

= - \frac{4}{9} \cdot \frac{ln ( \frac{14}{9} ) - ln ( \frac{4}{9} ) - ln ( \frac{10}{9} ) + ln ( \frac{8}{9} )}{2}

Simplify

= - \frac{2 ln ( \frac{14}{5} )}{9}

d er i v a t i v e f (x) = - 8 x^{3}

\int \frac{x ^{2} + 7 x + 1}{( x ^{2} + 1 ) ^{2}} d x

\frac{\partial}{\partial y} (2 x y^{4} - 3)

\int \frac{x + 2}{x ^{2} + 5 x - 6} d x

\int_{0}^{1} x^{2} x d x

What is the integral from 0 to 2 of 2/(y^2+y-20) ?
The integral from 0 to 2 of 2/(y^2+y-20) is -(2ln(14/5))/9