What is the integral from-2 to 3 of (38)/(x^4) ?

The integral from-2 to 3 of (38)/(x^4) is diverges

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integral from-2 to 3 of (38)/(x^4)

\int_{- 2}^{3} \frac{38}{x ^{4}} d x

diverges

Solution steps

\int_{- 2}^{3} \frac{38}{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{38}{x ^{4}} d x + \int_{0}^{3} \frac{38}{x ^{4}} d x

\int_{- 2}^{0} \frac{38}{x ^{4}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

x^{2} 4 x - 1

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

x \to - 2 lim (- x^{3} + x)

f (x) = \frac{x}{( 1 + x ^{2} )}, at x = 3

\int_{- 3}^{- 1} (x^{2} - 4 x + 8) d x

What is the integral from-2 to 3 of (38)/(x^4) ?
The integral from-2 to 3 of (38)/(x^4) is diverges