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

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

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integral from 0 to 1 of 6/(x^2+2x-8)

\int_{0}^{1} \frac{6}{x ^{2} + 2 x - 8} d x

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

Decimal

- 0.91629 \dots

Solution steps

\int_{0}^{1} \frac{6}{x ^{2} + 2 x - 8} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{1} \frac{1}{x ^{2} + 2 x - 8} d x

Factor

x^{2} + 2 x - 8 : - (- x^{2} - 2 x + 8)

= 6 \cdot \int_{0}^{1} \frac{1}{- ( - x ^{2} - 2 x + 8 )} d x

Take the constant out:

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

= 6 (- \int_{0}^{1} \frac{1}{- x ^{2} - 2 x + 8} d x)

Complete the square

- x^{2} - 2 x + 8 : - (x + 1)^{2} + 9

= 6 (- \int_{0}^{1} \frac{1}{- ( x + 1 ) ^{2} + 9} d x)

Apply u-substitution

= 6 (- \int_{1}^{2} \frac{1}{- u ^{2} + 9} d u)

Apply Integral Substitution

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

Take the constant out:

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

= 6 (- \frac{1}{3} \cdot \int_{\frac{1}{3}}^{\frac{2}{3}} \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}

= 6 (- \frac{1}{3} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{1}{3}}^{\frac{2}{3}})

Simplify

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

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

Compute the boundaries:

\frac{ln ( \frac{5}{3} ) + ln ( 3 ) - ln ( \frac{4}{3} ) + ln ( \frac{2}{3} )}{2}

= - 2 \cdot \frac{ln ( \frac{5}{3} ) + ln ( 3 ) - ln ( \frac{4}{3} ) + ln ( \frac{2}{3} )}{2}

Simplify

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

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

x \to 0 lim (\frac{( 3 + x ) - ( 3 )}{x})

12 y^{^{''}} - 5 y^{^{'}} - 2 y = 0

y^{^{'}} = 7 y^{2}

\frac{d}{d x} (\frac{x}{3 2 x})

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