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

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

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

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

diverges

Solution steps

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

Find undefined (singularity) points:

x = 1

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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\int_{1}^{2} \frac{1}{x} + y d x

\int_{8}^{10} 0.5 x^{- 1.5} d x

\int_{- 2}^{2} (1 + 2 x) d x

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