What is the integral from-1 to 3 of pi((15)/(x^4))^2 ?

The integral from-1 to 3 of pi((15)/(x^4))^2 is diverges

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integral from-1 to 3 of pi((15)/(x^4))^2

\int_{- 1}^{3} π (\frac{15}{x ^{4}})^{2} d x

diverges

Solution steps

\int_{- 1}^{3} π (\frac{15}{x ^{4}})^{2} 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_{- 1}^{0} π (\frac{15}{x ^{4}})^{2} d x + \int_{0}^{3} π (\frac{15}{x ^{4}})^{2} d x

\int_{- 1}^{0} π (\frac{15}{x ^{4}})^{2} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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What is the integral from-1 to 3 of pi((15)/(x^4))^2 ?
The integral from-1 to 3 of pi((15)/(x^4))^2 is diverges