What is the integral from-1 to 1 of (27)/(x^4)-3 ?

The integral from-1 to 1 of (27)/(x^4)-3 is diverges

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integral from-1 to 1 of (27)/(x^4)-3

\int_{- 1}^{1} \frac{27}{x ^{4}} - 3 d x

diverges

Solution steps

\int_{- 1}^{1} \frac{27}{x ^{4}} - 3 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} \frac{27}{x ^{4}} - 3 d x + \int_{0}^{1} \frac{27}{x ^{4}} - 3 d x

\int_{- 1}^{0} \frac{27}{x ^{4}} - 3 d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d}{d x} (a - b x)

\frac{\partial}{\partial t} (r cos (x))

y^{^{'}} = y (100 - y)

\int - \frac{sin ( x )}{cos ^{3} ( x )} d x

x \to 0 lim ((1 + 5 x)^{\frac{4}{x}})

What is the integral from-1 to 1 of (27)/(x^4)-3 ?
The integral from-1 to 1 of (27)/(x^4)-3 is diverges