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

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

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

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

diverges

Solution steps

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

\int_{- 2}^{0} \frac{1}{x ^{4}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{π}^{2 π} r sin (r) d r

\int_{\frac{- π}{2}}^{\frac{π}{2}} g (x) d x

\int_{0}^{1} (\frac{x - 1}{x + 1})^{3} d x

\int_{- 2}^{1} (3 x^{2} - 6 x + 2) d x

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

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