What is the integral from 0 to 1 of (x^3-5x-9)/(x^2-x-6) ?

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

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integral from 0 to 1 of (x^3-5x-9)/(x^2-x-6)

\int_{0}^{1} \frac{x ^{3} - 5 x - 9}{x ^{2} - x - 6} d x

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

Decimal Notation

1.82437 \dots

Solution steps

\int_{0}^{1} \frac{x ^{3} - 5 x - 9}{x ^{2} - x - 6} d x

Take the partial fraction of

\frac{x ^{3} - 5 x - 9}{x ^{2} - x - 6} : x + 1 + \frac{7}{5 ( x + 2 )} + \frac{3}{5 ( x - 3 )}

= \int_{0}^{1} x + 1 + \frac{7}{5 ( x + 2 )} + \frac{3}{5 ( x - 3 )} 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}{5 ( x + 2 )} d x + \int_{0}^{1} \frac{3}{5 ( x - 3 )} d x

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

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

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

\int_{0}^{1} \frac{3}{5 ( x - 3 )} d x = \frac{3}{5} (ln (2) - ln (3))

= \frac{1}{2} + 1 + \frac{7}{5} (ln (3) - ln (2)) + \frac{3}{5} (ln (2) - ln (3))

Simplify

\frac{1}{2} + 1 + \frac{7}{5} (ln (3) - ln (2)) + \frac{3}{5} (ln (2) - ln (3)) : \frac{3}{2} + \frac{- 4 ln ( 2 ) + 4 ln ( 3 )}{5}

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

\frac{1}{x - 2}

f (x) = 3 x^{3 x}

x \to 0 lim (\frac{1}{csc ( x )})

\frac{d}{d x} (\frac{2 x + 1}{3 x + 4})

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