What is the integral from 1 to 4 of 1/(p^2+p-6) ?

The integral from 1 to 4 of 1/(p^2+p-6) is diverges

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integral from 1 to 4 of 1/(p^2+p-6)

\int_{1}^{4} \frac{1}{p ^{2} + p - 6} d p

diverges

Solution steps

\int_{1}^{4} \frac{1}{p ^{2} + p - 6} d p

Find undefined (singularity) points:

p = 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{1}{p ^{2} + p - 6} d p + \int_{2}^{4} \frac{1}{p ^{2} + p - 6} d p

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{1}^{x} \frac{8}{3 t ^{3}} d t

\int_{0}^{1} 4 x^{3} - 2 x + 6 d x

\int_{1}^{6} 5 x - 5 d x

\int_{- 2}^{2} 6 4 - x^{2} d x

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

What is the integral from 1 to 4 of 1/(p^2+p-6) ?
The integral from 1 to 4 of 1/(p^2+p-6) is diverges