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

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

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

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

diverges

Solution steps

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d}{d t} (4 sin (3 t))

\int_{0}^{1} \frac{1}{x} d x

\int x cos (\frac{nπ x}{2}) d x

\frac{- u ^{2} + 3}{( u ^{2} + 3 ) ^{2}}

\int 2 cos (θ) cos^{5} (sin (θ)) d θ

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