What is the integral from 0 to 4 of 1/(x^2-x-2) ?

The integral from 0 to 4 of 1/(x^2-x-2) is diverges

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integral from 0 to 4 of 1/(x^2-x-2)

\int_{0}^{4} \frac{1}{x ^{2} - x - 2} d x

diverges

Solution steps

\int_{0}^{4} \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}^{4} \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

\frac{2}{s ( s ^{2} + 4 s + 3 )}

x \to \infty lim (\frac{6 x ^{2} + 2 x}{2 x ^{2} - 2 x + 1})

\int cos^{3} (\frac{x}{2}) d x

\frac{π r ^{2} ( π r ^{2} - 1 )}{4}

\frac{\partial}{\partial x} (x y e^{- x^{2} - y^{2}})

What is the integral from 0 to 4 of 1/(x^2-x-2) ?
The integral from 0 to 4 of 1/(x^2-x-2) is diverges