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

The integral from 1 to 3 of x/(x^2+x-6) is diverges

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

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

diverges

Solution steps

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{- 1}^{1} x^{3} + x^{5} + 7 x^{9} d x

\int_{0}^{16} x - 3 d x

\int_{1}^{3} (x^{2} - 6 x + 5) d x

\int_{2}^{4} 3 x - 1 + 2 x^{2} d x

\int_{0}^{2} 5 - t^{3} d t

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