What is the integral from 0 to 1 of (x^3)/(x^2-x-6) ?

The integral from 0 to 1 of (x^3)/(x^2-x-6) is 3/2-7/10 ln(9/4)+6*(2ln(2)-2ln(3))/5

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integral from 0 to 1 of (x^3)/(x^2-x-6)

\int_{0}^{1} \frac{x ^{3}}{x ^{2} - x - 6} d x

\frac{3}{2} - \frac{7}{10} ln (\frac{9}{4}) + 6 \cdot \frac{2 ln ( 2 ) - 2 ln ( 3 )}{5}

Decimal Notation

- 0.04076 \dots

Solution steps

\int_{0}^{1} \frac{x ^{3}}{x ^{2} - x - 6} d x

Factor

\frac{x ^{3}}{x ^{2} - x - 6}

= \int_{0}^{1} \frac{x ^{3}}{( x + 2 ) ( x - 3 )} d x

Long division

\frac{x ^{3}}{( x + 2 ) ( x - 3 )} : x + 1 + \frac{7 x + 6}{x ^{2} - x - 6}

= \int_{0}^{1} x + 1 + \frac{7 x + 6}{x ^{2} - x - 6} d x

Apply the Sum Rule:

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

= \int_{0}^{1} x d x + \int_{0}^{1} 1 d x + \int_{0}^{1} \frac{7 x + 6}{x ^{2} - x - 6} d x

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

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

\int_{0}^{1} \frac{7 x + 6}{x ^{2} - x - 6} d x = - \frac{7}{10} ln (\frac{9}{4}) + 6 \cdot \frac{2 ln ( 2 ) - 2 ln ( 3 )}{5}

= \frac{1}{2} + 1 - \frac{7}{10} ln (\frac{9}{4}) + 6 \cdot \frac{2 ln ( 2 ) - 2 ln ( 3 )}{5}

Combine the fractions

1 + \frac{1}{2} : \frac{3}{2}

= \frac{3}{2} - \frac{7}{10} ln (\frac{9}{4}) + 6 \cdot \frac{2 ln ( 2 ) - 2 ln ( 3 )}{5}

\frac{1}{s ^{2} + 40 s + 2000}

8 t^{2} y^{^{''}} - 16 y = 11 t^{2} - 1

cot (3 x)

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 6

What is the integral from 0 to 1 of (x^3)/(x^2-x-6) ?
The integral from 0 to 1 of (x^3)/(x^2-x-6) is 3/2-7/10 ln(9/4)+6*(2ln(2)-2ln(3))/5