What is the integral from-2 to 1 of (1/(x^2)-5) ?

The integral from-2 to 1 of (1/(x^2)-5) is diverges

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integral from-2 to 1 of (1/(x^2)-5)

\int_{- 2}^{1} (\frac{1}{x ^{2}} - 5) d x

diverges

Solution steps

\int_{- 2}^{1} \frac{1}{x ^{2}} - 5 d x

Find undefined (singularity) points:

x = 0

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_{- 2}^{0} \frac{1}{x ^{2}} - 5 d x + \int_{0}^{1} \frac{1}{x ^{2}} - 5 d x

\int_{- 2}^{0} \frac{1}{x ^{2}} - 5 d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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\int_{1}^{8} (\frac{x ^{2} + 3 x + 1}{x}) d x

\int_{2}^{7} \frac{x + 6}{x} d x

What is the integral from-2 to 1 of (1/(x^2)-5) ?
The integral from-2 to 1 of (1/(x^2)-5) is diverges