What is the integral from-13 to 13 of 1/(x^2+x-56) ?

The integral from-13 to 13 of 1/(x^2+x-56) is diverges

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integral from-13 to 13 of 1/(x^2+x-56)

\int_{- 13}^{13} \frac{1}{x ^{2} + x - 56} d x

diverges

Solution steps

\int_{- 13}^{13} \frac{1}{x ^{2} + x - 56} d x

Find undefined (singularity) points:

x = 7, x = - 8

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_{- 13}^{- 8} \frac{1}{x ^{2} + x - 56} d x + \int_{- 8}^{7} \frac{1}{x ^{2} + x - 56} d x + \int_{7}^{13} \frac{1}{x ^{2} + x - 56} d x

\int_{- 13}^{- 8} \frac{1}{x ^{2} + x - 56} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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\int_{0}^{0} f (t) sin (0 - t) d t

\int_{1}^{1024} \frac{e ^{5 x}}{5 x ^{4}} d x

What is the integral from-13 to 13 of 1/(x^2+x-56) ?
The integral from-13 to 13 of 1/(x^2+x-56) is diverges