What is the integral from 0 to 1 of 5/(x^2+x-6) ?

The integral from 0 to 1 of 5/(x^2+x-6) is -ln(8/3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of 5/(x^2+x-6)

\int_{0}^{1} \frac{5}{x ^{2} + x - 6} d x

- ln (\frac{8}{3})

Decimal Notation

- 0.98082 \dots

Solution steps

\int_{0}^{1} \frac{5}{x ^{2} + x - 6} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{1} \frac{1}{x ^{2} + x - 6} d x

Complete the square

x^{2} + x - 6 : (x + \frac{1}{2})^{2} - \frac{25}{4}

= 5 \cdot \int_{0}^{1} \frac{1}{( x + \frac{1}{2} ) ^{2} - \frac{25}{4}} d x

Apply u-substitution

: \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{4}{4 u ^{2} - 25} d u

= 5 \cdot \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{4}{4 u ^{2} - 25} d u

Take the constant out:

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

= 5 \cdot 4 \cdot \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{1}{4 u ^{2} - 25} d u

Apply Integral Substitution

= 5 \cdot 4 \cdot \int_{\frac{1}{5}}^{\frac{3}{5}} \frac{1}{10 ( v ^{2} - 1 )} d v

Take the constant out:

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

= 5 \cdot 4 \cdot \frac{1}{10} \cdot \int_{\frac{1}{5}}^{\frac{3}{5}} \frac{1}{v ^{2} - 1} d v

Factor

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

= 5 \cdot 4 \cdot \frac{1}{10} \cdot \int_{\frac{1}{5}}^{\frac{3}{5}} \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

= 5 \cdot 4 \cdot \frac{1}{10} (- \int_{\frac{1}{5}}^{\frac{3}{5}} \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}

= 5 \cdot 4 \cdot \frac{1}{10} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{1}{5}}^{\frac{3}{5}})

Simplify

5 \cdot 4 \cdot \frac{1}{10} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{1}{5}}^{\frac{3}{5}}) : - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{1}{5}}^{\frac{3}{5}}

= - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{1}{5}}^{\frac{3}{5}}

Compute the boundaries:

\frac{ln ( \frac{8}{5} ) - ln ( \frac{2}{5} ) - ln ( \frac{6}{5} ) + ln ( \frac{4}{5} )}{2}

= - 2 \cdot \frac{ln ( \frac{8}{5} ) - ln ( \frac{2}{5} ) - ln ( \frac{6}{5} ) + ln ( \frac{4}{5} )}{2}

Simplify

= - ln (\frac{8}{3})

\frac{( x + 5 ) ( x ^{2} - 5 x + 25 )}{x ^{3}}

\int (14 x + 5) 7 x^{2} + 5 x d x

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

\int \frac{1}{1 - x} d x

3 x (y^{2} + 1) d x + y (x^{2} + 2) d y = 0

What is the integral from 0 to 1 of 5/(x^2+x-6) ?
The integral from 0 to 1 of 5/(x^2+x-6) is -ln(8/3)