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

The integral from 0 to 3 of 1/(x^2-x-2) is diverges

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

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

diverges

Solution steps

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

Find undefined (singularity) points:

x = 2

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

x \to 3 lim (3 x^{2} - 2 x + 5)

\int \frac{sin ( x )}{cos ( x )} d x

\int 2 (3 x + 4)^{- 4} d x

f (x) = \frac{x}{x + 1}, (1, 2)

\frac{d}{d x} (\frac{x}{1 + sin ( x )})

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