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

The integral from-1 to 1 of x/(x^2+x-6) is 2(1/4 ln(2/3)+1/20 ln(6))

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

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

2 (\frac{1}{4} ln (\frac{2}{3}) + \frac{1}{20} ln (6))

Decimal Notation

- 0.02355 \dots

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

= \int_{- \frac{1}{2}}^{\frac{3}{2}} \frac{2 ( 2 u - 1 )}{4 u ^{2} - 25} d u

Take the constant out:

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

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int_{- \frac{1}{2}}^{\frac{3}{2}} \frac{2 u}{4 u ^{2} - 25} d u - \int_{- \frac{1}{2}}^{\frac{3}{2}} \frac{1}{4 u ^{2} - 25} d u)

\int_{- \frac{1}{2}}^{\frac{3}{2}} \frac{2 u}{4 u ^{2} - 25} d u = \frac{1}{4} ln (\frac{2}{3})

\int_{- \frac{1}{2}}^{\frac{3}{2}} \frac{1}{4 u ^{2} - 25} d u = - \frac{ln ( 6 )}{20}

= 2 (\frac{1}{4} ln (\frac{2}{3}) - (- \frac{ln ( 6 )}{20}))

Simplify

2 (\frac{1}{4} ln (\frac{2}{3}) - (- \frac{ln ( 6 )}{20})) : 2 (\frac{1}{4} ln (\frac{2}{3}) + \frac{1}{20} ln (6))

= 2 (\frac{1}{4} ln (\frac{2}{3}) + \frac{1}{20} ln (6))

\int \frac{x ^{2}}{256 - x ^{2}} d x

(tan^{2} (x))^{^{'}}

x \to 5 lim (\frac{( 13 - 4 x ^{2} + 69 )}{x - 5})

y = - x + 4, y = x^{2} + 3 x

\frac{d}{d x} (x e^{- 4 x})

What is the integral from-1 to 1 of x/(x^2+x-6) ?
The integral from-1 to 1 of x/(x^2+x-6) is 2(1/4 ln(2/3)+1/20 ln(6))